Probability, Class 12 Mathematics R.D Sharma Solutions

Page 32.13 Ex. 32.1

Q1.

Answer :

(i) P (X = 3) + P (X = 2) + P (X = 1) + P (X = 0) + P (X = -1)
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
=1.05 > 1
It is not the probability distribution of random variable X.

(ii) P (X = 0) + P (X = 1) + P (X = 2)
= 0.6 + 0.4 + 0.2
= 1.2 > 1
It is not the probability distribution of random variable X.

(iii) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
It is the probability distribution of random variable X.

(iv) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
= 0.3 + 0.2 + 0.4 + 0.1
= 1
It is the probability distribution of random variable X.

Q2.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = -2) + P (X = -1) + P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒0.1+k+0.2+2k+0.3+k=1⇒4k+0.6=1⇒k=0.44=0.1

Q3.

Answer :

(i) Since the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

⇒a+3a+5a+7a+9a+11a+13a+15a+17a=1⇒81a=1⇒a=181

(ii) P (X < 3)

=PX=0+PX=1+PX=2=181+381+581=981=19

P (X ≥ 3)

PX=3+PX=4+PX=5+PX=6+PX=7+PX=8=781+981+1181+1381+1581+1781=7281=89

P (0 < X < 5)
PX=1+PX=2+PX=3+PX=4=381+581+781+981=2481=827

Q4.

Answer :

(i) We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) = 1

⇒3c3+4c-10c2+5c-1=1⇒3c3-10c2+9c-2=0⇒c-13c2-7c+2=0⇒c-13c-1c-2=0⇒c= 13, 1, 2Neglecting 1 and 2 as individual probability should not be greater than one

(ii) P (X < 2)

=PX=0+PX=1=3c3+4c-10c2=19+43-109=1+12-109=39=13

(iii) P (1 < X ≤ 2)

=PX=2=5c-1=53-1=5-33=23

Q5.

Answer :

Let P (X = x3) = k. Then,
P (X = x₁) = k₂

P (X = x₂) = k₃

P (X = x₄) = k₅

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = x₁) + P (X = x₂) + P (X = x₃) + P (X = x₄) = 1

⇒k2+k3+k+k5=1⇒15k+10k+30k+6k30=1⇒61k30=1⇒k=3061

Now,

Q6.

Answer :

Let P (X = 0) = k. Then,
P (X = 0) = P (X > 0) = P (X < 0)
⇒P (X > 0) = k
P (X < 0) = k

∴ P (X = 0) + P (X > 0) + P (X < 0) = 1

⇒k+k+k=1⇒k=13

Now,
P (X < 0) = k

⇒PX=-1+PX=-2+PX=-3=k⇒3PX=-1=k ∵ PX=-1=PX=-2=PX=-3⇒PX=-1=k3⇒PX=-1=13×13=19∴ PX=-1=PX=-2=PX=-3=19Similarly, PX>0=k⇒PX=1=PX=2=PX=3=19

Thus, the probability distribution is given by

Q7.

Answer :

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=48C252C2=22562652=188221PX=1=P1 ace=4C1×48C152C2=1921326=32221PX=2=P2 aces=4C252C2=61326=1221

Thus, the probability distribution of X is given by

Q8.

Answer :

Let X denote the number of heads in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=PTTT=18, PX=1=PHTT or TTH or THT=38PX=2=PHTH or THH or HHT=38, PX=3=PHHH=18

Thus, the probability distribution of X is given by

Q9.

Answer :

Let X denote the number of aces in a sample of 4 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno ace=48C452C4PX=1=P1 ace=4C1×48C352C4PX=2=P2 aces=4C2×48C252C4PX=3=P3 aces=4C3×48C152C4PX=4=P4 aces=4C452C4

Thus, the probability distribution of X is given by

Page 32.14 Ex. 32.1

Q10.

Answer :

Let X denote the number of red balls in a sample of 3 balls drawn from a bag containing 4 red and 6 black balls. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno red ball=6C310C3=20120=16PX=1=P1 red ball=4C1×6C210C3=60120=12PX=2=P2 red balls=4C2×6C110C3=36120=310PX=3=P3 red balls=4C310C3=4120=130

Thus, the probability distribution of X is given by

Q11.

Answer :

Let X denote the number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective mango=15C420C4=13654845=91323PX=1=P1 defective mango=5C1×15C320C4=22754845=455969PX=2=P2 defective mangoes=5C2×15C220C4=10504845=70323PX=2=P3 defective mangoes=5C3×15C120C4=1504845=10323PX=3=P4 defective mangoes=5C420C4=54845=1969

Thus, the probability distribution of X is given by

Q12.

Answer :

Let X denote the sum of the numbers on two die. Then, X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Sample space : {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3 ,3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4 ,3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5 ,3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6 ,3), (6, 4), (6, 5), (6, 6)}

Now,
PX=2=136PX=3=236PX=4=336PX=5=436PX=6=536PX=7=636PX=8=536PX=9=436PX=10=336PX=11=236PX=12=136

Thus, the probability distribution of X is given by

Q13.

Answer :

Here, X can take the values 14, 15, 16, 17, 19, 20 and 21.
Now,
PX=14=215PX=15=115PX=16=215PX=17=315PX=18=115PX=19=215PX=20=315PX=21=115

Thus, the probability distribution of X is given by

Q14.

Answer :

Let X denote the number of defective bolts in a sample of 4 bolts drawn from a bag containing 5 defective bolts and 20 good bolts. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective bolts=20C425C4=484512650=9692530PX=1=P1 defective bolt=5C1×20C325C4=570012650=114253PX=2=P2 defective bolts=5C2×20C225C4=190012650=38253PX=3=P3 defective bolts=5C3×20C125C4=20012650=4253PX=4=P4 defective bolts=5C425C4=512650=12530

Thus, the probability distribution of X is given by

Q15.

Answer :

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=4852×4852=12×1213×13=144169PX=1=P1 ace=452×4852=2×1213×13=24169PX=2=P2 aces=452×452=1×113×13=1169

Thus, the probability distribution of X is given by

Q16.

Answer :

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno kings=4852×4852=12×1213×13=144169PX=1=P1 king=452×4852=2×1213×13=24169PX=2=P2 kings=452×452=1×113×13=1169

Thus, the probability distribution of X is given by

Q17.

Answer :

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=4852×4751=22562652=188221PX=1=P1 ace=452×4851+4852×451=3842652=32221PX=2=P2 aces=452×351=122652=1221

Thus, the probability distribution of X is given by

Q18.

Answer :

Let X denote the number of white balls in a sample of 3 balls drawn from a bag containing 4 white and 6 red balls. Then, X can take the values 0, 1, 2 and 3.
Now,

PX=0=Pno white ball=6C310C3=20120=16PX=1=P1 white ball=4C1×6C210C3=60120=12PX=2=P2 white balls=4C2×6C110C3=36120=310PX=3=P3 white balls=4C310C3=4120=130

Thus, the probability distribution of X is given by

Q19.

Answer :

It is given that Y denotes the number of times a total of 9 appears on throwing the pair of dice.

When the dice is thrown 2 times, the possibility of getting a total of 9 is possible only for the given combinations:

(3, 6) (4, 5) (5, 4) (6, 3)

So, the total number of outcomes is 36 and the total number of favourable outcomes is 4.

Probability of getting a total of 9 = 436=19

Probability of not getting a total of 9 = 1-19=89

If Y takes the values 0, 1 and 2, then

PY=0=89×89=6481PY=1=19×89+89×19=1681PY=2=19×19=181

Thus, the probability distribution of X is given by

Q20.

Answer :

Let X denote the number of defective items in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective item=20C425C4=484512650=9692530PX=1=P1 defective item=5C1×20C325C4=570012650=114253PX=2=P2 defective items=5C2×20C225C4=190012650=38253PX=3=P3 defective items=5C3×20C125C4=20012650=4253PX=4=P4 defective items=5C425C4=512650=12530

Thus, the probability distribution of X is given by

Q21.

Answer :

Let X denote the number of hearts in a sample of 3 cards drawn from a well-shuffled deck of 52 cards. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno heart=3952×3952×3952=2764PX=1=P1 heart=1352×3952×3952+3952×1352×3952+3952×3952×1352=2764PX=2=P2 hearts=1352×1352×3952+3952×1352×1352+1352×3952×1352=964PX=3=P3 hearts=1352×1352×1352=164

Thus, the probability distribution of X is given by

Q22.

Answer :

Let X denote the number of blue balls in a sample of 3 balls drawn from a bag containing 4 red and 3 blue balls. Then, X can take values 0, 1, 2 and 3.
Now,
PX=0=Pno blue ball=47×47×47=64343PX=1=P1 blue ball=37×47×47+47×37×47+47×47×37=144343PX=2=P2 blue balls=37×37×47+47×37×37+37×47×37=108343PX=3=P3 blue balls=37×37×37=27343

Thus, the probability distribution of X is given by

Q23.

Answer :

Let X denote the number of spades in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

PX=0=Pno spade=39C252C2=7411326=1934PX=1=P1 spade=13C1×39C152C2=5071326=1334PX=2=P2 spades=13C252C2=781326=117

Thus, the probability distribution of X is given by

Q24.

Answer :

Let X denote the event of getting a number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
PX=0=1636=49PX=1=1636=49 PX=2=436=19

Thus, the probability distribution of X is given by

Q25.

Answer :

The possible values of X are 0, 1 and 2, i.e. no black ball, 1 black ball and 2 black balls.
Yes, X is a random variable.
A random variable is a real valued function having domain as the sample space associated with a random experiment.

Q26.

Answer :

Given: X = Number of heads − Number of tails

Therefore, the possible values of X are :
−6, −4, −2, 0, 2, 4, 6

Q27.

Answer :

Let X denote the number of defective bulbs in a sample of 2 bulbs drawn from a lot of 10 bulbs containing 3 defectives and 7 non-defectives.Then X can take values 0, 1, 2.
Now,
PX=0=Pno defective bulb=7C210C2=715PX=1=P1 defective bulb=3C1×7C110C2=715PX=2=P2 defective bulbs=3C210C2=115

Thus, the probability distribution of X is given below,

Page 32.34 Ex. 32.2

Q2.

Answer :

Let X denote the number of heads in three tosses of a coin.Then, X can take values 0, 1, 2 and 3.
Now,
PX=0=PTTT=18, PX=1=PHTT or TTH or THT=38PX=2=PHTH or THH or HHT=38, PX=3=PHHH=18

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=32Variance=∑pixi2-Mean2 =3-322 =34

Q3.

Answer :

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno king=48C252C2=11281326=188221PX=1=P1 king=4C1×48C152C2=1921326=32221PX=2=P2 kings=4C252C2=61326=1221

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=34221Variance=∑pixi2-Mean2            =36221-342212          =7956-115648841          =680048841          =4002873

Q4.

Answer :

Let X denote the number of tails in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=PHHH=18, PX=1=PTHH or HHT or HTH=38PX=2=PTTH or THT or HTT=38, PX=3=PTTT=18

Thus, the probability distribution of X is given by

Computation of mean and step deviation

Mean=∑pixi=32Variance=∑pixi2-Mean2  =3-322 =34Step Deviation=Variance =34    =0.87

Q5.

Answer :

Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno bad egg=10C312C3=120220=611PX=1=P1 bad egg=2C1×10C212C3=90220=922PX=2=P2 bad eggs=2C2×10C112C3=10220=122

Thus, the probability distribution of X is given by

Computation of mean

Mean=∑pixi=12

Q6.

Answer :

Let X denote the event of getting twice the number. Then, X can take the values 1, 2, 3, 4, 5 and 6.

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2.5Variance=∑pixi2-Mean2=8.4-6.25=2.15

Q7.

Answer :

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTT, HTTT, THHH, …

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when a coin is tossed 4 times, we can get minimum 0 and maximum 4 strings.)

Now,
PX=0=P0 head=116PX=1=P1 head=416PX=2=P2 heads=616PX=3=P3 heads=416PX=4=P4 heads=116

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2Variance=∑pixi2-Mean2  =5-4  =1

Q8.

Answer :

Let X denote the event of getting twice the number. Then, X can take the values 2, 4, 6, 8, 10 and 12.

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=7Variance=∑pixi2-Mean2  =60.7-49 =11.7

Q9.

Answer :

Let X be 1 for the appearance of odd numbers 1, 3 or 5 on the die. Then,
PX=1=36=12

Let X be 3 for the appearance of even numbers 2, 4 or 6 on the die. Then,
PX=3=36=12

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2Variance=∑pixi2-Mean2  =5-4  =1

Q10.

Answer :

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH and TTTT .

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when the coin is tossed 4 times, we can get maximum 4 and minimum 0 strings.)

Now,
PX=0=P0 head=116PX=1=P1 head=716PX=2=P2 heads=516PX=3=P3 heads=216PX=4=P4 heads=116

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2716=1.7Variance=∑pixi2-Mean2   =6116-729256  =247256 =0.9

Page 32.36 (Very Short Answers)

Q1.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = -2) + P (X = -1) + P (X = 0) + P (X = 1) = 1

⇒1-a4+1+2a4+1-2a4+1+a4=1⇒44=1⇒1=1Now, 0≤1-a4≤1⇒ 0≤1-a≤4⇒ -1≤-a≤3⇒ -3≤a≤1 …10≤1+a4≤1⇒ 0≤1+a≤4⇒ -1≤a≤3 …20≤1-2a4≤1⇒ 0≤1-2a≤4⇒ -1≤-2a≤3⇒ -32≤a≤12 …30≤1+2a4≤1⇒ 0≤1+2a≤4⇒ -1≤2a≤3⇒ -12≤a≤32 …4From 1, 2, 3 and 4, we get-12≤a≤12

Q2.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒2k4+3k2-5k3+2k-3k2+3k-1=1⇒2k4-5k3+5k=2⇒2k4-5k3+5k-2=0⇒k-1k-22k2+k-1=0⇒k-1k-22k-1k+1=0⇒k=-1 , 12, 1, 2Neglecting -1 , 1 and 2 as they give the value of probability negative or greater than 1

∴ k =12

Q3.

Answer :

A cubical die can show 1, 2, 3, 4, 5 or 6 on its face.

Mean = ∑pixi = 16+26+36+46+56+66=216=3.5

Q4.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒2k+4k+3k+k=1⇒10k=1⇒k=110=0.1

Q5.

Answer :

Mean = ∑pixi = 14+28+158=2+2+158=198

Q6.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒c+2c+4c+4c=1⇒11c=1⇒c=111Now,PX≤2=PX=1+PX=2=110+210=311

Page 32.37 (Very Short Answers)

Q7.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒k+2k+3k+4k=1⇒10k=1⇒k=110Now, PX≥3=PX=3+PX=4=310+410=710

Page 32.37 (Multiple Choice Questions)

Q1.

Answer :

(d) 181
We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

⇒a+3a+5a+7a+9a+11a+13a+15a+17a=1⇒81a=1⇒a=181

Q2.

Answer :

(b) 0.77

E = {X is a prime number} = {2, 3, 5, 7}

PE =P2+P3+P5+P7= 0.62F= {X<4} = {X=1,2,3} PF= P1+P2+P3=0.5Now,E∩F = {2,3}PE∩F=P2+P3=0.35PE∪F=PE+PF-PE∩F=0.62+0.50-0.35=0.77

Q3.

Answer :

(d) 0.4

Let:
P(X = 0) = m
P(X = 1) = k.

Now,
P(X = 3) = 2k

Mean = ∑pixi

0+k+0.6+6k=1.3⇒7k=1.3-0.6⇒k=0.77=0.1

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒m+0.1+0.3+0.2=1⇒m+0.6=1⇒m=0.4

Q4.

Answer :

(a) 1/10

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

⇒0+2p+2p+3p+p2+2p2+7p2+2p=1⇒10p2+9p-1=0⇒10p-1p+1=0⇒p=110 or -1 Neglecting -1 as the value of the probabilitiy cannot be negative

Q5.

Answer :

(d) 1/8, 3/4

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒k+3k+3k+k=1⇒8k=1⇒k=18

Now,

Mean=∑pixi=32Variance=∑pixi2-Mean2=248-322=248-94=24-188=68=34

Page 33.12 Ex. 33.1

Q1.

Answer :

Let X denote the number of defective items in a sample of 8 items. Then, X follows a binomial distribution with n = 8, p = (probability of getting a defective item) = 0.06 and q = 1-p=0.94

P(X=r) =Cr8 (0.06)r (0.94)8-r, r=0,1,2,3,…8The required probability = Probability of not more than one defective item = P (X≤1) = P(X=0)+P(X=1) = C08 (0.06)0 (0.94)8-0+ C18 (0.06)1 (0.94)8-1 = (0.94)8 +8(0.06)(0.94)7 = (0.94)7 0.94+0.48 = 1.42(0.94)7

Q2.

Answer :

Let X denote the number of heads in 5 tosses.X follows a binomial distribution with n =5; p = probability of getting a head = 12and q = 1-p =12P(X=r) = Cr512r125-r, r =0,1,2…5The required probability = P(getting at least 3 heads)= P(X≥3) = P(X=3)+P(X=4)+P(X=5) = C35123125-3+ C45121125-1 + C55125125-0
=10125 + 5125 + 1125 = 125(10+5+1) = 125×16 = 12

Q3.

Answer :

Let X denote the number of tails when a coin is tossed 5 times.X follows a binomial distribution with n =5; p = 12; q =1-p =12Then P(X=r) =Cr512r 12n-r = Cr5125 The required probability = P(X=odd) =P(X=1)+P(X=3)+P(X=5)=C15125 +C35125 +C55125 = 125 [5+10+1] = 1632=12

Q4.

Answer :

Let X be the number of successes in 6 throws of the two dice.

Probability of success = Probability of getting a total of 9
= Probability of getting (3,6), (4,5), (5,4), (6,3) out of 36 outcomes

p=436=19, q =1-p =89and n = 6X follows a binomial distribution with n=6, p=19 and q=89P(X=r) = Cr619r 896-r The required probability = Probability of at least 5 successes= P(X≥5) = P(X=5)+P(X=6)=C56 195 896-5 +C66 196 896-6 = 6(8)+196=4996

Q5.

Answer :

Let X denote the number of heads when 6 coins are tossed.Then, X follows a binomial distribution with n =6; p = 12and q =12;P(X=r) = Cr612r126-r = Cr6125, r =0,1,2…..6

The required probability= P( X≥3) = P(X=3)+P(X=4)+P(X=5)+P(X=6)= C36126+ C46126+ C56126+ C66126=(20 + 15++6+1)164=2132

Q6.

Answer :

Let X be the probability of getting 4 in two tosses of a fair die.
X follows a binomial distribution with n =2; p =16 and q =56;
P(X=r) = Cr216r562-rProbability of getting 4 at least once = P(X≥1) = 1-P(X=0) = 1- C02160562-0 = 1-2536= 1136

Q7.

Answer :

Let X be the number of heads that appear when a coin is tossed 5 times.

X follows a binomial distribution with n =5 and p= q=12
P(X=r) = Cr512r125-r = Cr5125

P (head appears an even number of times)=P(X=0)+P(X=2)+P(X=4)= C05125+C25125+C45125 = 1+10+525= 1632=12

Q8.

Answer :

Let X be number of times the target is hit. Then, X follows a binomial distribution with n =7, p =14and q =34

P(X=r) = Cr714r347-rP( hitting the target at least twice)=P(X≥2) = 1-P(X=0) +P(X=1)= 1-C07140347-0-C17141347-1 = 1-347 – 714346= 1-116384(2187+5103) = 1-36458192=45478192

Q9.

Answer :

Let X be the number of busy calls for 6 randomly selected telephone numbers.

X follows a binomial distribution with n =6 ;p = one out of 15 = 115and q = 1415
P(X=r) = Cr6115r14156-rProbability that at least 3 of them are busy = P(X≥3) = 1-{P(X=0)+P(X=1)+P(X=2)}= 1 -C06115014156-0+C16115114156-1+C26115214156-2

=1-14156+61514155+11514154

Q10.

Answer :

Let X denote the number of successes, i.e. of getting 5 or 6 in a throw of die in 6 throws.

Then, X follows a binomial distribution with n =6;
p = of getting 5 or 6 = 16+16=13; q = 1-p = 23;P(X=r) = Cr613r236-rP(X ≥4) = P(X=4)+P(X=5)+P(X=6)=C46134236-4+C56135236-5+C66136236-6 = 136(60+12+1) =73729

Q11.

Answer :

Let X be the number of heads in tossing 8 coins.
X follows a binomial distribution with n =8; p = 12and q = 12;
P(X=r) = Cr812r128-r = Cr8128Probability of obtaining at least 6 heads = P(X≥6) = P(X=6)+P(X=7)+P(X=8)= C68128 + C78128 + C88128 = 12828+8+1 = 37256

Q12.

Answer :

Let X denote the number of spade cards when 5 cards are drawn with replacement. Because it is with replacement,

X follows a binomial distribution with n = 5; p =1352=14; q = 1-p = 34

P(X=r) =Cr514r345-r

(i) P(All cards are spades) = P(X=5) =C55145340 = 11024(ii) P(only 3 cards are spades) = P(X=3) =C35143342 = 1102490 = 45512iii P(none is a spade) = P(X=0)=C05140345 = 2431024

Q13.

Answer :

Let X be the number of white balls drawn when 4 balls are drawn with replacement.
X follows binomial distribution with n = 4.

p = Probability for a white ball = No of white ballsTotal no. of balls = 520=14and q = 1-p = 34P(X=r) = Cr414r344-r(i) Prob that none is white = P(X=0) =C04140344-0= 81256ii Prob that all are white = P(X=4) = C44144344-4= 1256iii Prob that any two are white = P(X=2) = C24142344-2= 54256=27128

Page 33.13 Ex. 33.1

Q14.

Answer :

Let X be the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.
Then, X follows a binomial distribution with n =5;

p = Probability of getting a ticket bearing number divisible by 10. p = 1100(10) = 110; q=910;P(X=r) = Cr5110r9105-rProbability that all the tickets bear numbers divisible by 10= P(X=5) =C5511059105-5=11059100=1105

Hence, required probability is 1105

Q15.

Answer :

Let X be the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

As this is with replacement, X follows a binomial distribution with n = 4;
p = probabilty that a ball randomly drawn bears digit 0 =110;q=1-p = 910;

P(X=r) = Cr4110r9104-r P(none bears the digit 0) = P(X=0) = C0411009104-0 = 9104

Q16.

Answer :

Let X denote the number of defective items in a sample of 10 items.

X follows a binomial distribution with n =10; p = probability of defective items = 5% = 0.05 ;q = 1-p =0.95

P(X=r) =Cr10(0.05)r(0.95)10-rProbability ( sample of 10 items will include not more than one defective item)= P(X≤1) = P(X=0)+P(X=1) =C010(0.05)0(0.95)10-0+C110(0.05)1(0.95)10-1 =( 0.95)9(0.95+0.5)= 1.45(0.95)9

=192092920; (∵ 1.45 = 2920and 0.95 = 1920)

Q17.

Answer :

Let X be the number of bulbs that fuse after 150 days.

X follows a binomial distribution with n = 5, p =0.05 and q =0.95
Or p = 120and q = 1920P(X=r) = Cr5120r19205-r(i) Probability (none will fuse after 150 days of use ) = P(X=0) =C05120019205-0 =19205(ii) Probability (not more than 1 will fuse after 150 days of use) = P(X≤1) = P(X=0)+P(X=1) =19205+5C1120119205-1 = 192041920+520 = 6519204
(iii) Probability more than one will fuse after 150 days of use = P(X>1) = 1-P(X≤1)= 1-6519204 ∵P(X≤1)=6519204 iv Probability at least one will fuse after 150 days of use = P(X≥1)= 1-P(X=0) = 1-19205 ∵P(X=0=19205

Q18.

Answer :

Let X be the number of people that are right-handed in the sample of 10 people.

X follows a binomial distribution with n = 10,
p =90% =0.9 and q = 1-p = 0.1P(X=r) = Cr10(0.9)r(0.1)10-rProbability that at most 6 are right-handed = P(X≤6)= P(X=0)+ P(X=1)+ P(X=2)+ P(X=3)+ P(X=4)+ P(X=5)+ P(X=6)= 1-{P(X=7)+ P(X=8)+ P(X=9)+ P(X=10)}= 1- ∑r=710Cr10(0.9)r(0.1)10-r

Q19.

Answer :

Let X denote the number of red balls drawn from 16 balls with replacement.
X follows a binomial distribution with n = 4,p = 516, q =1-p = 1116

P(X=r) = Cr4516r11164-rP(One ball is red)=P(X=1) = C14516111164-1= 451611163=5411163

Q20.

Answer :

Let X denote the number of white balls when 2 balls are drawn from the bag.

X follows a distribution with values 0,1 or 2.
P(X=0) = P(All balls non-white) = C27C29= 4272=2136P(X=1) = P ( Ist ball white and IInd ball non-white )  =C17C12C29= 1436P(X=2) = P(Both balls white) = C22C29= 136It can be shown in tabular form as follows.X 0 1 2P(X) 2136 1436 136

Q21.

Answer :

As three balls are drawn with replacement, the number of white balls, say X, follows binomial distribution with n =3
p =37 and q =47P(X=r) = Cr337r473-r, r=0,1,2,3P(X=0) =C03370473-0 P(X=1) = C13371473-1 P(X=2) = C23372473-2 P(X=3) =C33373473-3X 0 1 2 3P(X) 64343 144343 108343 27343

Q22.

Answer :

Let X be the number of doublets in 4 throws of a pair of dice.

X follows a binomial distribution with n =4,
p = No of getting (1,1)(2,2)…(6,6) = 636=16q =1-p = 56P(X=r) = Cr416r564-r, r= 0,1,2,3,4P(X=0) = C04160564-0P(X=1) = C14161564-1P(X=2) = C24162564-2P(X=3) = C34163564-3P(X=4) = C44164564-4The distribution is as follows.X 0 1 2 3 4 P(X) 6251296 5001296 1501296 201296 11296

Q23.

Answer :

Let X be the number of 6 in 3 tosses of a die.

Then X follows a binomial distribution with n =3.

p = 16, q=1-p =56P(X=r) = Cr316r563-r, r=0,1,2,3P(X=0) = C03160563-0P(X=1) = C13161563-1P(X=2) = C23162563-2P(X=3) = C33163563-3

Hence, the distribution of X is as follows.X 0 1 2 3P(X) 125216 75216 15216 1216

Q24.

Answer :

Let X = number of heads in 5 tosses. Then the binomial distribution for X has n =5, p =12 and q =12
P(X=r) = Cr512r125-r, r = 0,1,2,3,4,5 = Cr525Substituting r =0,1,2,3,4,5 we get the following probability disrtribution.X 0 1 2 3 4 5P(X) 132 532 1032 1032 532 132

Q25.

Answer :

Let X denote getting a number greater than 4 .

Then, X follows a binomial distribution with n=2

p = P(X>4)=P(X=5 or 6)= 16+16=13q =1-p = 23P(X=r) = Cr2 13r232-r , r=0,1,2 Substituting for r we get probability distribution of X as follows.

X 0 1 2P(X) 49 49 19

Q26.

Answer :

Let X be the number of rupees the man wins.

First, let us assume that he gets head in the first toss.Probability would be 12. Also, he wins Re.1.The second possibility is that he gets a tail in the first toss. Then he tosses again.Suppose he gets a head in the second toss.Then, he wins Re 1 in the second toss but loses Re 1 in the first toss. So, the money won =Rs. 0Probability for winning Rs .0 =Getting tail in Ist toss x Getting head in IInd toss=12×12=14The third possibility is getting tail in the Ist toss and also tail in the IInd toss.Then, the money that he would win = -2 (As he loses Rs. 2)Probability for the third possibility = 12×12=14Probability distribution of X is as follows.X 1 0 -2P(X) 12 1 4 14

Q27.

Answer :

Let X denote the occurrence of 3,4 or 5 in a single die. Then, X follows binomial distribution with
n=5.
Let p=probability of getting 3,4, or 5 in a single die .

p = 36=12

q=1-12=12 P(X=r) = Cr512r125-rP(at least 3 successes) =P(X≥3) = P(X=3)+P(X=4)+P(X=5)= C35123125-3+C45124125-4+C55125125-5= C35+C45+C5525= 12

Q28.

Answer :

Let X denote the number of defective items in the items produced by the company.
Then, X follows binomial distribution with n = 8.

p = 10%=110q= 1-p = 910Hence, the distribution is given byP(X=r) = Cr8110r9108-rProb of getting 2 defective items =P(X=2) =C2811029108-2 = 28 x 96108

Q29.

Answer :

(i) Let p denote the probability of drawing a heart from a deck of 52 cards. So,

p=1352=14and q=1-q=1-14 = 34

Let the card be drawn n times. So, binomial distribution is given by: P(X=r)=Crnprqn-r
Let X denote the number of hearts drawn from a pack of 52 cards.
We have to find the smallest value of n for which P(X=0) is less than 14
P(X=0) < 14
C0n14034n-0<1434n<14Put n=1, 341 not less than 14 n=2, 342 not less than 14 n=3, 343 not less than 14So, smallest value of n=3
Therefore card must be drawn three times.

(ii) Given the probability of drawing a heart > 34
1 – P(X=0) > 34
1- C0n14034n-0>341-34n>341-34>34n14>34n

For n=1, 341 not less than 14 n=2, 342 not less than 14 n=3, 343 not less than 14 n=4, 344 not less than 14 n=5, 345 not less than 14
So, card must be drawn 5 times.

Q30.

Answer :

Let k be the number of desks and X be the number of graduate assistants in the office.
therefore, X=8, p=12, q=12
According to the given condition,
PX≤k>90%⇒PX≤k>0.90⇒PX>k<0.10⇒P(X=k+1, k+2,….8)<0.10
Therefore, P(X > 6) = P(X=7 or X=8)
C78128 + C88128 =0.04

Now, P(X > 5) = P(X = 6, X = 7 or X = 8) = 0.15
P(X > 6) < 0.10

So, if there are 6 desks then there is at least 90% chance for every graduate to get a desk.

Q31.

Answer :

Let X be the number of heads in tossing the coin 8 times.
X follows a binomial distribution with n =8
p =12 and q=12Hence, the distribution is given by ∴P(X=r) = Cr812r128-r, r=0,1,2,3,4,5,6,7,8Required probability =P(X≥6) = P(X=6)+P(X=7)+P(X=8)= C68+ C78+ C8828= 28+8+1256=37256

Q32.

Answer :

Let X denote the number of heads obtained in tossing 6 coins.
Then, X follows a binomial distribution with n=6,

p = 12 and q=12Hence, the distribution is given byP(X=r) =Cr612r126-r , r=0,1,2,3,4,5,6= Cr626i P(getting 3 heads) =P(X=3) =C3626= 2064=516ii P(getting no head) =P(X=0) =C0626= 126 = 164iii P(getting at least 1 head) = P(X≥1) =1- P(X=0) = 1-164=6364

Page 33.14 Ex. 33.1

Q33.

Answer :

Let X denote the number of tubes that function for more than 500 hours.
Then, X follows a binomial distribution with n =4.
Let p be the probability that the tubes function more than 500 hours.

Here, p=0.2, q =0.8Hence, the distribution is given byP(X=r) =Cr4(0.2)r(0.8)4-r, r= 0,1,2,3,4Therefore, required probability = P(X=3) = 4(0.2)3(0.8) = 0.0256

Q34.

Answer :

Let X denote the number of components that survive shock.
Then, X follows a binomial distribution with n = 5.
Let p be the probability that a certain kind of component will survive a given shock test.
∴p = 34and q=14Hence, the disrtibution is given byP(X=r) = Cr534r145-r, r=0,1,2,3,4,5i P(exactly 2 will survive)=P(X=2) =C25342145-2 = 10×91024= 0.0879

ii P(atmost 3 will survive) = P(X≤3)= P(X=0)+ P(X=1)+ P(X=2) +P(X=3)=C05340145-0+C15341145-1+C25342145-2+C35343145-3= 145+534144+10342143+10343142= 1+15+90+2701024= 3761024= 0.3672

Q35.

Answer :

Let X be the number of bombs that hit the target.
Then, X follows a binomial distribution with n = 6
Let p be the probability that a bomb dropped from an aeroplane will strike the target.

∴ p =0.2 and q =0.8Hence, the distribution is given byP(X=r) =Cr60.2r0.86-r(i) P(exactly 2 will strike the target)=P(X=2) = C26(0.2)2(0.8)4=15(0.04)0.4096=0.2458(ii) P(at least 2 will strike the target) = P(X≥2) = 1-[P(X=0)+P(X=1)]= 1- (0.8)6-6(0.2)(0.8)5 = 1-0.2621-0.3932= 0.3447

Q36.

Answer :

Let X be the number of mice that contract the disease .
Then, X follows a binomial distribution with n =5.
Let p be the probability of mice that contract the disease.
∴ p=0.4 and q = 0.6Hence, the distribution is given byP(X=r)=Cr50.4r0.65-r , r=0,1,2,3,4,5i P(X=0) =C050.400.65-0= (0.6)5 = 0.0778ii P(X>3) =P(X=4)+P(X=5)=C450.440.65-4+C550.450.65-5=0.0768+0.01024= 0.08704

Q37.

Answer :

Let X denote the number of successes in 6 trials.

It is given that successes are twice the failures. ⇒p=2q p+q=1⇒3q=1⇒q=13∴ p=1-13=23 n =6Hence, the distribution is given by

P(X=r)=Cr623r136-r , r=0,1,2…..6P(atleast 4 successes)=P(X≥4) = P(X=4)+P(X=5)+P(X=6)C46234136-4+C56235136-5+C66236136-6= 15(24)+6(32)+6436= 240+192+64729= 496729

Q38.

Answer :

Let X denote the number of machines out of service during a day.Then, X follows a binomial distribution with n=20 Let p be the probability of any machine out of service during a day.∴p= 0.02 and q =0.98 Hence, the distribution is given byP(X=r) =Cr20 0.02r0.9820-r , r=0,1,2…..20∴P(exactlly 3 machines will be out of the service on the same day)=P(X=3) =C3 200.0230.9820-3 = 1140(0.000008)(0.7093)= 0.006469

Q39.

Answer :

Let X be the number of students that graduate from among 3 students.
Let p=probability that a student entering a university will graduate.

Here , n =3, p=0.4 and q = 0.6
Hence, the distribution is given by
P(X=r)=Cr30.4r0.63-r , r=0,1,2,3

i P(X=0) = q3 =0.216ii P(X=1) = 3(0.4)(0.36) =0.432iii P(X =3) = p3 = 0.064

Q40.

Answer :

Let X be the number of defective eggs drawn from 10 eggs.
Then, X follows a binomial distribution with n =10
Let p be the probability that a drawn egg is defective.

∴p =10%=110 , q =910Hence, the distribution is given by P(X=r) =Cr10110r91010-r , r=0,1,2….10P(there is at least one defective egg) =P(X≥1) = 1-P(X=0) =1-C010110091010-0= 1-91010

=1-9101010

Q41.

Answer :

Let X denote the number of correct answers.
Then, X follows a binomial distribution with n =20

Let p be the probability of a correct answer. ⇒p= getting a head and a right answer to be true or getting a tail and a right answer to be false⇒p= 12∴ q=1- 12=12Hence, the distribution is given byP(X=r) = Cr2012r1220-r , r=0,1,2,3……20= Cr20220Probability that the student answers at least 12 questions correctly=P(X≥12) = P(X=12)+P(X=13)+…+P(X=20)= C1220+C1320+…+C2020220

Q42.

Answer :

We have n=6 and p=12∴ q = 1-p =12Hence, the distribution is given byP(X=r) =Cr612r126-r, r=0,1,2,3,4,5,6= Cr6126

P(X=r) = Cr626 By substituting r =0,1,2,3,4,5 and 6, we get the following distribution for X.X 0 1 2 3 4 5 6P(X) 164 664 1564 2064 1564 664 164

Comparing the probabilities, we get that X = 3 is the most likely outcome.

Q43.

Answer :

Let X be the number of right answers in the 5 questions.
X can take values 0,1,2…5.
X follows a binomial distribution with n =5
.p =probability of guessing right answer = 13 q =probability of guessing wrong answer = 23Hence, the distribution is given byP(X=r)=Cr5 13 r 235-r , r=0,1,2,…5∴ P(The student guesses 4 or more correct answers) = P(X≥4) = P(X=4)+P(X=5)=C45134231+C55135230 = 10+135=11243

Q44.

Answer :

Let X denote the number of times the person wins the lottery.
Then, X follows a binomial distribution with n = 50.

Let p be the probability of winning a prize.∴ p = 1100, q=1-1100=99100Hence, the distribution is given byP(X=r) = Cr501100r9910050-r , r=0,1,2 …50i P(winning at least once) = P(X≥0) = 1-P(X-0)= 1-9910050
ii P(winning exactly once) = P(X=1) =C150110019910050-1 = 129910049

iii P(winning at lease twice) = P(X≥2) = 1-P(X=0)-P(X=1)= 1-9910050-C150 ×1100×9910049 =1-9949 ×14910050

Q45.

Answer :

Let the shooter fire n times and let X denote the number of times the shooter hits the target.
Then, X follows binomial distribution with p =34and q =14 such that

PX=r=Crn 34r14n-r ⇒PX=r=Crn 3r4nIt is given that PX≥1 >0.99⇒1-PX=0 >0.99⇒1-14n>0.99⇒14n<0.01⇒4n>10.01⇒4n>100The least value of n satisfying this inequality is 4. Hence, the shooter must fire at least 4 times.

Q46.

Answer :

Suppose the man tosses a fair coin n times and X denotes the number of heads in n tosses.
As p=12 and q=12, P(X=r) =Crn12r12n-r, r=0,1,2,3….nIt is given that PX≥1 >0.9⇒1-PX=0 >0.9⇒1-C0n 12n>0.9⇒12n<110⇒2n>10⇒n=4,5,6….Hence, the man must toss the coin at least 4 times.

Q47.

Answer :

Let X be the number of heads and n be the minimum number of times that a man must toss a fair coin so that probability of X≥1 is more than 80% and X follows a binomial distribution with

p = 12, q=12P(X=r) = Crn12nWe have P(X≥1) = 1-P(X=0) =1-C0n12n =1-12nand P(X≥1) >80%1-12n>80% = 0.8012n<1-0.80 =0.20 2n>10.2 =5; We know, 22 <5 while 23>5So, n =3 So, n should be atleast 3.

Q48.

Answer :

Let p denote the probability of getting a doublet in a single throw of a pair of dice. Then,
p = 636 = 16q = 1-p=1-16=56

p= Let X be the number of getting doublets in 4 throws of a pair of dice. Then, X follows a binomial distribution with n =4,
p=16and q=56P(X=r)=Probability of getting r doubletsP(X=r) = Cr4(16)r(56)4-r; r= 0,1,2,3,4If X=0, then P(X=0) = C04(16)0(56)4-r⇒P = 564If X=1, then P = C14161564-1=23563If X=2, then P = C24162564-2=16562 If X=3, then P = C34163564-3=103163

If X=4, then P =C44(16)4(56)4-4 =164= 11296

X 0 1 2 3 4P(X) (56)4 23(56)3 25216 5324 11296

Q49.

Answer :

Let X denote the number of defective bulbs in a sample of 4 bulbs drawn successively with replacement .
Then, X follows a binomial distribution with the following parameters: n=4, p =630=15and q =45
Then, the distribution is given byP(X=r) =Cr415r454-r , r=0,1,2,3,4

P X=0=454 = 256625 P X=1 = 4151453 = 256625 PX=2= 6152452 = 96625 PX=3 = 4153451 = 16625 PX=4 =154 = 1625

X 0 1 2 3 4

P(X) 256625 256625 96625 16625 1625

Page 33.24 Ex. 33.2

Q1.

Answer :

No.

The mean of a binomial distribution is np and variance is npq.

If mean is less than its variance, then np <npq.

As both n and p are positive, we can divide both sides by np.

We get 1<q, which is not true as q <1 under all circumstances. (As p+q=1, q cannot be greater than 1)

So, the mean of a binomial distribution cannot be less than its variance.

Q2.

Answer :

It is given that mean, i.e. np = 9 and variance, i.e. npq =94

∴npqnp =94×19 = 14 Hence, q=14 and p =1-q = 34When p = 34,np = 3n4 9=3n4⇒n =12P(X=r) =Cr1234r1412-r, r=0,1,2,3,4,5,…..12

Q3.

Answer :

Given: Mean = 9 and variance = 6

∴ np =9 …(1) npq =6 …(2) Dividing eq (2) by eq (1), we get q =23and p =1-q=13As np =9, substituting the value of p, we get n3=9 or n =27P(X=r) = Cr2713r2327-r, r=0,1,2….27

Q4.

Answer :

Number of trials in the binomial distribution = 5

If p is the probability for success, then

np + npq = 4.8
Or 5p+5p (1-p) = 4.8

⇒5p + 5p-5p2 =4.8 Or p2-2p+0.96 =0By factorising, we get (p-0.8)(p-1.2) =0As p cannot exceed 1, p =0.8 or 45and q= 1-p = 15∴ P(X=r) = Cr545r155-r, r=0,1,2,….5

Q5.

Answer :

Mean, i.e. np =20 ….(1)

Variance, i.e. npq =16 ….(2)

Dividing eq (2) by eq (1), we get npqnp= 1620⇒q = 45⇒ p = 1-q ∴p= 15 As np = 20 ⇒ n = 100∴P(X=r) =Cr10015r45100-r , r=0,1,2….100

Q6.

Answer :

Given:Sum of the mean and variance= 253⇒np+npq=253 ⇒np 1+q=253 …(1)Product of the mean and variance = 503 ⇒np(npq) = 503 …(2)Dividing eq (2) by eq (1), we get np(npq)np(1+q)= 503×325⇒npq1+q=2⇒npq = 2(1+q) ⇒np(1-p) = 2(2-p)⇒np = 2(2-p)(1-p) Substituting this value in np+npq =253, we get2(2-p)(1-p) (2-p) = 253⇒ 6(4-4p+p2) = 25-25p⇒6p2+p-1 =0 ⇒(3p-1)(2p+1) =0 ⇒p =13or -12. As p cannot be negative, the only answer for p is 13.
q=1-p =23 ⇒np+npq = 253⇒n131+23 =253⇒n = 15∴P(X=r) = Cr1513r2315-r, r=0,1,2……15

Q7.

Answer :

Given that mean, i.e. np = 20 …(1)
and standard deviation, i.e. npq = 4

npq =4 ⇒npq =16 …(2)Dividing eq (2) by eq (1), we get q= 1620=45and p = 15; ∴ n = Meanp= 100 P(X=r) = Cr10015r45100-r, r=0,1,2…..100Therefore, the parameters are n=100 and p = 15

Q8.

Answer :

Total number of bolts (n) = 400 and p = prob of defective bolts = 0.1

(i) Mean = np = 400(0.1) =40

(ii) Variance = npq = 40(1-0.1) = 36

So, the standard deviation = 36 =6

Q9.

Answer :

Mean of binomial distribution, i.e. np =5

Variance, i.e. npq =103
q= VarianceMean= 23and p = 1-q= 13np = 5 ⇒ n= 15∴P(X=r) = Cr1513r2315-r, r=0,1,2……15

Q10.

Answer :

Total number of ships (n) = 500
Let X denote the number of ships returning safely to the ports.
p=910 and q =1-p =110

Mean = np = 450 and Variance =npq = 45 Mean =450 Standard deviation = 45 = 6.71

Q11.

Answer :

Given: mean =16 and variance = 8
Let n and p be the parameters of the distribution.

That is, np = 16 and npq = 8

q= npqnp=12and p = 1-q =12 np = 16 ⇒n = 32∴P(X=r) = Cr3212r1232-r, r=0,1,2….32,⇒P(X=0) =1232 P(X=1) = 321232=122 P(X≥2) = 1-P(X=0)-P(X=1)= 1-1232-1227= 1-1+32232 = 1-33232

Q12.

Answer :

Let X denote the number of successes in 8 throws.

n =8
p = probability of getting 5 or 6 =26=13 and q = 23

Mean (np) =83Variance (npq) = 169Standard deviation= Variance = 43So, mean = 2.66 and standard deviation = 1.33

Q13.

Answer :

Here, n =8
Let p be the probability of number of boys in the family.

p =12, q=12Expected number of boys =mean ⇒ np = 4

Q14.

Answer :

Here, n =10,000
Let p (the probability of getting a defective item) = 0.02
q =1-0.02 = 0.98

Mean= Expected number of defective items ⇒np =200Variance (npq) = 200(0.98) = 196Standard deviation = Variance = 14So, mean =200 and standard deviation =14

Q15.

Answer :

Here, n =3
p = probability of getting 1 or 6 =13and q =1-13=23Mean = np =1Variance = npq = 23

Q16.

Answer :

Mean(np) =3 and variance (npq)=32∴ q=12and p =1-12n=Meanp ⇒n=6

Hence, the distribution is given by P(X=r) = Cr612r126-r , r=0,1,2…6= Cr626 ∴P(X≤5) = 1-P(X=6) = 1-164=6364

Q17.

Answer :

Here, mean (np) = 4
variance (npq ) =2

∴q= 12and p =12 n=Mean p=4×2=8
Hence, the distribution is given by P(X=r) =Cr812r128-r, r=0,1,2……8P(X≥5) = P(X=5)+P(X=6)+P(X=7)+P(X=8)=128C58+C68+C78+C88 = (56+28+8+1)28= 93256

Page 33.25 Ex. 33.2

Q18.

Answer :

Mean np =43and Variance npq = 89∴q = 23and p =1-23= 13Therefore, n = Meanp = 4
Hence, the distribution is given by P(X=r) = Cr413r234-r, r=0,1,2,3,4P(X≥1) = 1-[P(X=0)] = 1-234 = 81-1681= 6581

Q19.

Answer :

Given that n = 6
The sum of mean and variance of a binomial distribution for 6 trials is 103.
⇒6p+6pq = 103⇒18p+18p(1-p) =10 ⇒18p2-36p+10 =0⇒(3p-1)(6p-10) =0 ⇒ p =13or 53p=53 (Neglected as it is greater than 1)∴ p=13⇒q= 1-p = 23Hence, the distribution is given by P(X=r) = Cr613r236-r, r=0,1,2…..6

Q20.

Answer :

Let X be the number of times a doublet is obtained in four throws.

Then, p = probability of success in one throw of a pair of dice =636=16
and q = 56; n = 4P(X=r) = Cr416r564-r, r= 0,1,2,3,4As n =4 and p = 16, mean = np = 46= 23
∴ P(X=r) = Cr456r16n-r, r=0,1,2,3,4The distribution is as follows:X 0 1 2 3 4P(X) 564 2064 15064 50064 164

Q21.

Answer :

Total number of outcomes when two dice are thrown = 6 × 6, i.e. 36

Let X be the number of doublets in three throws of a pair of dice.
Then, X follows a binomial distribution with n = 3
p = P(Getting a doublet in three throws) =636=16and q = 56∴P(X=r) =Cr316r563-r, r=0,1,2,3Mean np = 316 = 12
The distribution is as follows:X 0 1 2 3P(X) 563 3161563-1 3162563-2 163 125216 75216 15216 1216

Page 33.25 (Very Short Answers)

Q1.

Answer :

n= 20 , q =0.75

⇒p = 1-q = 0.25Mean = np = 20(0.25) =5Thus, mean =5

Q2.

Answer :
Mean =5 and Variance= 4⇒np =5 and npq =4 ⇒ q= 0.8 ⇒p =1-q =0.2& np = n(0.2) =5 (given)⇒n = 50.2= 25

Q3.

Answer :
It is given that the binomial distribution’s p =0.2 and number of items (n) = 200

Hence, mean, i.e. np = 200 (0.2) =40

Q4.

Answer :

Standard deviation of the binomial distribution =4 Variance=Standard deviation2Variance, i.e. npq= 16 Mean =np = 20⇒q = 1620=45and p = 1-q = 15

Page 33.26 (Very Short Answers)

Q5.

Answer :

Mean of the binomial distribution, i.e. np =10Variance = Standard deviation2, i.e. npq =4∴q= VarianceMean =0.4

Q6.

Answer :

Given, np=4 and npq=2∴p = 1-VarianceMean = 1-24 =12and q=12 and n = npp =412 = 8Hence, the binomial distribution is given byPX=r=Cr8 12r128-r , r=0,1,2,3…….8∴P(X=1) = 8128 = 132

Q7.

Answer :
Mean =2 ,Variance =1∴ q=VarianceMean= 12and p =1-12=12n = Meanp= 212=4The binomial distribution is given byP(X=r)=Cr412r124-r∴ P(X=0)=C04120124-0 , r=0,1,2,3,4=124P(X>1) = 1-P(X=0) = 1-124 = 1516

Q8.

Answer :

In the given binomial distribution, n = 4 and
P(X=0) = 1681 Binomial distribution is given byP(X=0)=C04 p0q4-0=q4 We know that P(X=0) = 1681 ∴ q4 =1681⇒ q4=234⇒q =23

Q9.

Answer :

Mean (np)= 4

Variance (npq )= 3

.⇒q = 34 Hence, p =1-34 = 14 and n = Meanp= 4×4 =16Therefore, the binomial distribution is given byP(X=r) =Cr16 14r3416-r , r=0,1,2….rProbability of no success = C0161403416-0=3416

Q10.

Answer :

For binomial distribution of X,P(X=r)= Crn(p)r(q)n-r, r= 0,1,2,…, nP(X=1)=np(q)n-1 P(X=2) =C2np2 (q)n-2⇒np(q)n-1=C2np2 (q)n-2 =α Simplifying the above equation we get,q=n-12p ⇒2q = np-p On putting, q =1-p we get 2-2p = np-p p(n+1) =2 ….. (i)Also, P(X=1) = α⇒np(1-p)n-1 =α …..(ii)

Note: We cannot find the value of n as (i) and (ii) are not linear and hence we cannot find the value of P(X = 4)

Page 33.26 (Multiple Choice Questions)

Q1.

Answer :

(a) 9105

Let X denote the number of defective bulbs.

Hence, the binomial distribution is given by

n =5 , p = 10100=110 & q = 90100=910Hence, the distribution is given by P(X=r)=Cr5110r9105-r∴P(X=0)= 9105

Q2.

Answer :

(b) 181

In the given binomial distribution, n = 4 and
P(X=0) = 1681 Binomial distribution is given byP(X=0)=C04 p0q4-0=q4 We know that P(X=0) = 1681 ∴ q4 =1681⇒ q4=234⇒q =23∴ p=1-23=13Then , P(X=4)=C44 p4q4-4=134=181

Q3.

Answer :

(c) 7

Let p=chance of hitting a distant target
⇒p =10% or p= 0.1

⇒q =1-0.1= 0.9Let n be the least number of rounds .P(hitting atleast once) = P(X≥1) ⇒ 1-P(X=0) ≥50%⇒ 1-P(X=0) ≥ 0.5P(X=0) ≤0.5⇒(0.9)n≤0.5Taking log on both the sides, we get n log 0.9 ≤ log 0.5 ⇒n≤log 0.5log 0.9⇒n≤7.2 Therefore, 7 is the least number of rounds that he must fire in order to have more than 50% chance of hitting the target at least once.

Q4.

Answer :

(c) 15/2¹³

Let X denote the number of heads in a fixed number of tosses of a coin .Then, X is a binomial variate
with parameters n and p=12

Given that P (X=7) =P (X = 9). Also, p = q = 0.5

P(X=r) = Crn(0.5)r(0.5)n-r =Crn (0.5)n∴P(X=7) =C7n(0.5)n and P(X=9)=C9n(0.5)nIt is given that P(X=7)=P(X=9)∴ C7n(0.5)n =C9n(0.5)n⇒n!7! n-7 !=n!9! n-9 !⇒9×8=n-7n-8⇒n2-8n-7n+56=72⇒n2-15n-16=0⇒n+1n-16=0⇒n=-1 or n=16 ⇒n=- 1 (Not possible as n denotes the number of tosses of a coin)∴ n=16 Hence, P(X=2) =C216(0.5)16 = 16.152×1216= 15213

Q5.

Answer :

(a) 1/2

Here n=100
Let X denote the number of times a tail is obtained.

Here, p =q = 12P(X=odd) = P(X=1,3,5,….99) = C1100+C3100+…..+C9910012100= Sum of odd coefficients in binomial expansion in (1+x)10012100= 2(100-1)2100= 12

Q6.

Answer :

(c) 84×56610
Let p be the probabilty of obtaining a six in a single throw of the die .Then ,p =16and q =1-16=56Obtaining a fourth six in the tenth throw of the die means that in the first nine throws there are 3 sixes and the fourth six is obtained in the tenth throw. Therefore, required probability=P(Getting 3 sixes in the first nine throws) P(Getting a six in the tenth throw)=C39 p3q9-3 p=C39163566×16= 84 x 56610

Q7.

Answer :

(a) 1/2

Given that P(X=r) = k P(X=n-r), where k is independent of n and r .

Crn prqn-r = k Cn-rn pn-rqrWe have Crn=Cn-rn and also q=1-pHence, the equation changes to the following:pr(1-p)n-r = k pn-r(1-p)r⇒(1-p)n-2r = k pn-2r⇒qpn-2r=k This is possible when p =q and k becomes 1.Hence, p = q = 12

Q8.

Answer :

(a) 7, 14

Here, p=12and q=12Binomial distribution is given by P(X=r)=Crn12r12n-r

P (X = 4), P (X = 5), P(X = 6) are in A.P.

∴C4n+C6n=2C5n⇒n(n-1)(n-2)(n-3)24!+n(n-1)(n-2)(n-3)(n-4)(n-5)26!= n(n-1)(n-2)(n-3)(n-4)5!By simplifying, we get12+(n-4)(n-5)2(30)= n-45Taking LCM as 60, we get30+n2-9n+20 = 12n-48 ⇒n2-21n+98 =0 ⇒(n-7)(n-14) =0⇒n =7, 14

Page 33.27(Multiple Choice Questions)

Q9.

Answer :

(b) 51/101

Let X denote the number of coins showing head. Therefore, X follows a binomial distribution with p and n as parameters.Given that P(X=50)=P(X=51)⇒C50100p50q50 = C51100p51q49on simplifying we get,5150=pq⇒5150=p1-p (since p+q=1)⇒p=51101

Q10.

Answer :

(a) 49, 50

When a coin is tossed 99 times, the number of heads X follows a binomial distribution with
p = q=12= 0.5P(X=r) =Crn(0.5)r(0.5)n-r=Crn(0.5)nAs (0.5)n is common to all r it is enough if we find the maximum of Crn.We know that for odd number of n, there will be two equal maximum terms,i.e. when r=n-12and r = n+12Hence, n =99 So, the maximum is obtained when r = 49 or 50

Q11.

Answer :

(d) 3

Let X denote the number of coins.
Then, X follows a binomial distribution with

p = 12 , q = 12It is given that P(X≥1) ≥0.8⇒1-P(X=0) ≥0.8⇒P(X=0)≤1-0.8 ⇒P(X=0)=0.2⇒12n≤0.2 ⇒2n≥10.2⇒2n≥5This is possible when n≥3So, the least value of n is 3.

Q12.

Answer :

(d) 15/16

Mean =2 and variance =1

⇒np =2 and npq = 1⇒q=12 ⇒p = 1-12=12 n=Meanp⇒n =4Hence, the distribution is given byPX=r=Cr412r124-r , r=0,1,2,3,4∴P(X≥1) = 1-P(X=0) = 1-124=1516

Q13.

Answer :

(a) 1/3

Probability of heads = p
Probability of Tails = (1 − p)
Probability that first head appears at even turn
Pe = (1 − p)p + (1 − p)3p+(1 − p)5p + …..
= (1 − p)p (1 + (1 − p)2 + (1 − p)4+ …..)

=1-pp11-1-p2=1-pp1-p2+2p

Pe=1-p2-pPe=251-p2-p=255-5p=4-2p3p=1p=13Q14.

Q14.

Answer :

(b) 119128

n=8, p = 12∴ q=1-12=12Hence, the distribution is given byP(X=r) = Cr812r128-r PX-4≤2 = P(-2≤X-4≤2) =P(2≤X≤6)=P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=C28+C38+C48+C58+C6828= 28+56+70+56+28256=238256=119128

Q15.

Answer :

(c) 33

The binomial distribution is given by,
fr, n, p=PX=r=crn pr 1-pn-r
This value can be maximum at a particular r, which can be determined as follows,
fr+1, n, pfr, n, p=1⇒cr+1npr+1×1-pn-r-1crnpr×1-pn-r=n-rpr+1 1-p=1
On substituting the values of n = 100, p=13, we get
100-r×13r+1 1-13=1⇒100-r13=r+123⇒100-r=r+12⇒100-r=2r+2⇒98=3r⇒3r=98∴r=983
The integer value of r satisfies (n + 1)p − 1 ≤ m < (n + 1)p

f (r, n, p) is montonically increasing for r < m and montonically decreasing for r > m

as  983≤m<1013

∴ The integer value of r is 33.

Q16.

Answer :

(b) C27×5568
Let p be the probabilty of obtaining a six in a single throw of the die .Then ,p =16and q =1-16=56Obtaining a third six in the eighth throw of the die means that in first seven throws there are 2 sixes and the third six is obtained in the eighth throw . Therefore,required probability=P(Getting 2 sixes in the first seven throws) P(Getting six in the eighth throw)=C27 p2q7-2 p=C27162565 ×16= C27 x 5568

Q17.

Answer :

Answer: (d)

Let p= probability that a selected coupon bears number ≤9.
p=915=35
n = number of coupons drawn with replacement
X = number of coupons bearing number ≤9
Probability that the largest number on the selected coupons does not exceed 9
= probability that all the coupons bear number ≤9
= P(X=7) = C77p7q0=377
Similarly, probability that largest number on the selected coupon bears the number ≤8 will be

P(X=7) = C77p7q0=8157 (since, p will become 815)
Hence required probability will be = 377-8157
So, option (d)

Q18.

Answer :

Let number be abcde
Case 1 : e = 0
a, b, c can be filled in 9 × 9 × 9 ways
c = 0 ⇒ 9 × 8 × 1 ways and d has 9 choices
c ≠ 0 ⇒ (9 × 9 × 9 – 9 × 8 × 1) = 657
in the case d has 8 choices ⇒ 657 × 8
Total case = 9 × 8 × 1 × 9 + 657 × 8 ⇒ 5904
Case 2 : e = 5
If c = 5,
if a ≠ 5 then a, b, c can be filled in 8 × 8 × 1 = 64 ways
if a = 5 then a, b, c can be filled in 1 × 9 × 1 = 9 ways
if c ≠ 5, then first 3 digits can be filled in 729 – 64 – 9 = 656 ways
here d has 8 choices
No. of member ending in 5 and no two consecutive digits being identical ⇒ (64 + 9) × 9 + 656 × 8
⇒ 5905
Total cases ⇒ 5904 + 5905 ⇒ 11809

Q19.

Answer :

(b) 105512

Let X denote the number of heads obtained in 10 tosses of a coin . Then, X follows a binomial distribution with n =6 , p=12=qThe distribution is given by P(X=r)=Cr1012r1210-r∴ P(X=6) = C610210= 10529=105512

Q20.

Answer :

(b) C6161463410

Mean (np) = 4 and Variance (npq) = 3

∴ q=34⇒p =1-34=14and n = 16Let X denotes the number of successes in 16 trials .Then, P(X=r) =Cr1614r3416-r⇒P(X=6) = Probability (getting exactly 6 successes)= 16C61463410

Q21.

Answer :

(c) 12

p =14∴ q= 1-14=34Standard deviation =3 ⇒Variance = 32=9Or npq = 9⇒ np =9q= 363=12Therefore, the mean of the given binomial distribution is 12.

Page 33.28(Multiple Choice Questions)

Q22.

Answer :

(d) 1516

Let X denote the number of heads obtained in four tosses of a coin .
Then X follows a binomial distribution with
n =4 and p =q =12Distribution is given by P(X=r)=Cr412r124-r ∴P(X=r)=C04120124-0 P(atleast one head turns up) = P(X≥1) = 1-P(X=0) = 1-124=1516

Q23.

Answer :

n =3

P(X=1) =8 P(X=3) (Given)The distribution is given by P(X=r)=Cr3prq3-rP(X=1)=C13p1q2 and P(X=3)=C33p3q0⇒3pq2 = 8p3 ⇒8p2 = 3q2 ⇒8p2= 3(1-p)2 ⇒8p2= 3-6p+3p2⇒5p2+6p-3=0⇒p=-6±9610

Hence , it does not match any of the answer choices.

Q24.

Answer :

(b) 3

Let X be the number of heads. Then X follows a binomial distribution with

p =12, q=12Hence, the distribution is given by P(X=r) = Crn12r12n-r, r=0,1,2,3……n∴P(X≥1) =1-P(X=0) = 1-12n ≥0.8Or 2n≥10.2⇒ 2n≥5This is possible only when n ≥3.So, the least value of n must be 3.

Q25.

Answer :

(d) 12

Let X be the number of males.
p = q = 12 (given)P(X=n-1) =Cn-1npn-1q1 = 3210⇒n12n = 3210 ⇒n12n = 322212 ⇒n12n= 121212By comparing the two sides, we get n =12