12th NCERT Probability Exercise 13.4 Questions 17
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Question (1)

State which of the following are not the probability distributions of a random variable. Give reasons for your answer. (i)
x 0 1 2
p(x) 0.4 0.4 0.2

Solution

∑ p(x) = 0.4 + 0.4 + 0.2
= 1.
Yes it is probability distribution function.
(ii)
x 0 1 2 3 4
P(x) 0.1 0.5 0.2 -0.1 0.3

Solution

As one of the probability is negative , it is not probability distribution function.
(iii)
x -1 0 1
p(x) 0.6 0.1 0.2

Solution

∑p(x) = 0.6 + 0.1+ 0.2 = 0.9 < 1
As the sum is less than 1, so no it is not probability distribution function.
(iv)
z 3 2 1 0 -1
p(x) 0.3 0.2 0.4 0.1 0.05

Solution

∑ p(x) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 > 1
As sum of probabilities is greater than 1, so it is not probability distribution function.

Question (2)

An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

Solution

in an urn 5 red + 2 black = 7 balls
Two balls drawn at random. n = 7C2 = 21.
X ; the number of black balls
X = 0, 1, 2.
\[P\left( {x = 0} \right) = \frac{{5{C_2}}}{{7{C_2}}} = \frac{{5 \times 4}}{{7 \times 6}} = \frac{{10}}{{21}}\] \[P\left( {x = 1} \right) = \frac{{5{C_1} \times 2{C_1}}}{{21}} = \frac{{5 \times 2}}{{21}} = \frac{{10}}{{21}}\] \[P\left( {x = 2} \right) = \frac{{2{C_2}}}{{21}} = \frac{1}{{21}}\] \[\sum {p(x) = \frac{{10}}{{21}} + \frac{{10}}{{21}} + \frac{1}{{21}} = 1} \] So yes it is probability distribution function.

Question (3)

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Solution

X ; difference between number of heads and number of tails when coin is tossed 6 times.
No. of heads No. of tails Difference.
0 6 6
1 5 4
2 4 2
3 3 0
4 2 2
5 1 4
6 0 6

So possible values of X are 0, 2, 4, and 6.

Question (4)

Find the probability distribution of (i) number of heads in two tosses of a coin (ii) number of tails in the simultaneous tosses of three coins (iii) number of heads in four tosses of a coin

Solution

(i) The coin is tossed two times.
So sample space S = { HH, HT, TH, TT}. n = 4
X ; number of heads in two tosses of a coin.
X = 0, 1, 2.
P(X = 0) = 1/4., P(X = 1) = 2/4, P(X = 2) = 1/4.
The probability distribution function is as follows.
X 0 1 2 Total
p(X) 1/4 2/4 1/4 1

(ii) The coin is tossed three times.
So sample space S = { HHH, HHT,HTH, HTT, THH,THT, TTH TTT}. n = 8
X ; number of heads in three tosses of a coin.
X = 0, 1, 2, 3.
P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8.
The probability distribution function is as follows.
X 0 1 2 3 Total
p(X) 1/8 3/8 3/8 1/8 1

(iii) The coin is tossed four times.
So sample space S = { HHHH, HHHT,HHTH, HHTT, HTHH,HTHT, HTTH HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}. n = 16
X ; number of heads in four tosses of a coin.
X = 0, 1, 2, 3, 4.
P(X = 0) = 1/16, P(X = 1) = 4/16, P(X = 2) = 6/16, P(X = 3) = 4/16, P(X = 4) = 1/16.
The probability distribution function is as follows.
X 0 1 2 3 4 Total
p(X) 1/16 4/16 6/16 4/16 1/16 1

Question (5)

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 (ii) six appears on at least one die

Solution

The die is thrown two times. so sample space will contain 36 elements. n = 36.
(i) The success is defined as number greater than 4.
X ; the number of sucess
X = 0, 1, 2.
p = probability of seccess = 2/6
\[P\left( {x = 0} \right) = \frac{4}{6} \times \frac{4}{6} = \frac{{16}}{{36}}\] \[P\left( {x = 1} \right) = \frac{4}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{4}{6} = \frac{{8 + 8}}{{36}} = \frac{{16}}{{36}}\] \[P\left( {x = 2} \right) = \frac{2}{6} \times \frac{2}{6} = \frac{4}{{36}}\] The probability distribution function is given as follows.
X 0 1 2 Total
P(X) 16/36 16/36 4/36 1

(ii) Six appears on at least one die.
X = 0, 1.
P(X = 0) = 25/36, P(X = 1) = 11/36.
X 0 1 Total
P(X) 25/36 11/36 1

Question (6)

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Solution

6 bulbs out of 30 bulbs are selected with replacement.
The probability of selected bulb is defective = 6/30 = 1/5.
The probability that selected bulb is not defective = 24/ 30 = 4/5.
X = Number of defective bulbs
X = 0, 1, 2, 3, 4.
\[P\left( {x = 0} \right) = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = \frac{{256}}{{625}}\] \[P\left( {x = 1} \right) = 4C1 \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5} = 4 \times \frac{{64}}{{125}} \times \frac{1}{5} = \frac{{256}}{{625}}\] \[P\left( {x = 2} \right) = 4{C_2} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5} \times \frac{1}{5} = 6 \times \frac{{16}}{{25}} \times \frac{1}{{25}} = \frac{{96}}{{625}}\] \[P\left( {x = 3} \right) = 4{C_3} \times \frac{4}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = 4 \times \frac{4}{5} \times \frac{1}{{125}} = \frac{{16}}{{625}}\] \[P\left( {x = 4} \right) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{{625}}\] The probability distribution function is given as follows.
X 0 1 2 3 4 Total
p(X) 256/625 256/625 96/625 16/625 1/625 1

Question (7)

A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Solution

P(H) = 3P(T)
P(H) + P(T) = 1
3P(T) + P(T) = 1
4P(T) = 1
P(T) = 1/4, P(H) = 3/4.
A coin is tossed twice , so sample space S = {HH, HT, TH, TT}
X ; the number of tails.
X = 0, 1, 2.
P(X = 0 ) = P( HH)
\[P\left( {x = 0} \right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{{16}}\] P(X = 1) = P( HT, TH)
\[P\left( {x = 1} \right) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{{3 + 3}}{{16}} = \frac{6}{{16}}\] P(X = 2) = P(TT)
\[P\left( {x = 2} \right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\]The probability distribution function is as follows
X 0 1 2 Total
P(X) 9/16 6/16 1/16 16/16=1

Question (8)

. A random variable X has the following probability distribution:
X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k
Determine (i) k (ii) P(X < 3) (iii) P(X > 6) (iv) P(0 < X < 3)

Solution

(i) As it is given the probability distribution function,
∑p(x) = 1
0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
10k2 + 9k = 1
10 k2 + 9k - 1 = 0
( k + 1)(10k - 1 ) = 0
k + 1 = 0 or 10k - 1 = 0
k = -1 or k = 1/10
As k is the probability k ≠ -1. so k = 1/10.
(ii) P(X < 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
=3k
= 3/10
(iii) P( X > 6) = P(7)
= 7k2 + k
= 7 (1/10)2 + 1/10
= 7/100 + 1/10
= 17/100
(iv) P(0 < X < 3) = P(1) + P(2)
= k + 2k
= 3k
= 3/10.

Question (9)

The random variable X has probability distribution P(X) of the following form, where k is some number: P(X) = { 𝑘, if x = 0 2𝑘, if x = 1 3𝑘, if x = 2 0, otherwise (a) Determine the value of k. (b) Find P(X < 2),P( X≤2) , P(X ≥ 2),

Solution

( a ) The probability distribution function for the random variale is given.
∑p(x) = 1
k + 2k + 3k +0 = 1
6k = 1
k = 1/6
( b ) P( X < 2 ) = P( x = 0) + P( x = 1)
= k + 2k
= 3k
= 3(1/6) = 1/2
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= k + 2k + 3k
= 6k
=6(1/6) = 1
P(X ≥ 2) = P(X = 2)
= 3k
=3(1/6) = 1/2.

Question (10)

Find the mean number of heads in three tosses of a fair coin.

Solution

As coin is toss three times , n = 8
X ; the number of heads.
X = 0, 1, 2, 3.
P(X = 0) = 1/8, P(X = 1) = P(X = 2) = 3/8, P(X = 3) = 1/8.
X P(X) X P(X)
0 1/8 0
1 3/8 3/8
2 3/8 6/8;
3 1/8 3/8
Total 1 12/8

Mean of X = E(X)
= ∑ XP(X)
= 12/8
= 1.5

Question (11)

Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Solution

Two dics are thrown , so n = 36.
X ; the number of sixes.
X = 0, 1, 2.
P(X = 0) = 25/36, P(X = 1) = 10/36, P(X = 2) = 1/36
X P(X) X P(X)
0 25/36 0
1 10/36 10/36
2 1/36 2/36
Total 36/36 = 1 12/36;

E(X) = ∑XP(X)
= 12/36
= 1/3.

Question (12)

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denotes the larger of the two numbers obtained. Find E(X).

Solution

There are 6 positive numbers , Two number are selected with out replacement.
S = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6)}
n = 15.
X ; the large number between them
X = 2, 3, 4, 5, 6.
X P(X) X P(X)
2 1/15 2/15
3 2/15 6/15
4 3/15 12/15
5 4/15 20/15
6 5/15 30/15
Total 15/15=1 70/15

E(X) = ∑XP(X)
= 70/15
= 14/3

Question (13)

Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X

Solution

When two dice are rolled then n = 36.
X ; Sum og the numbers on dics.
X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
X P(X) XP(X) X2P(X)
2 1/36 2/36 4/36
3 2/36 6/36 18/36
4 3/36/td> 12/36 48/36
5 4/36 20/36 100/36
6 5/36 30/36 180/36
7 6/36 42/36 294/36
8 5/36 40/36; 320/36
9 4/36 36/36 324/36
10 3/36 30/36 300/36
11 2/36 22/36 242/36
12 1/36 12/36 144/36
Total 36/36 = 1 /252/36 1974/36

E(X) = ∑XP(X)
= 252 / 36
= 7
E(X2) = ∑X2P(X)
= 1974/ 36
=54.833
V(X) = E(X2) - [E(X)]2
= 54.833 - (7)2
= 54.833 - 49
= 5.833
S.D = √V(X)
= √5.833
= 2.415.

Question (14)

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Solution

X ; age of the student
X = 14,15, 16, 17, 18, 19, 20, 21.
X frquency P(X) XP(X) X2P(X)
14 2 2/15 28/15 392/15
15 1 1/15 15/15 225/15
16 2 2/15 32/15 512/15
17 3 3/15 51/15 867/15
18 1 1/15 18/15 324/15
19 2 2/15 38/15 722/15
20 3 3/15 60/15 1200/15
21 1 1/15 21/15 441/15
Total 15 1 263/15 4683/15

E(X) = ∑XP(X)
= 263/15
=17.53
E(X2) = ∑X2P(X)
= 4683/15
V(X) = E(X2) - [E(X)]2
= 4683/ 15 - (263/15)2
= (70245 - 69169)/225
= 1076 / 225
=4.78
S.D.(X) = √V(X)
= √4.78
=2.186 = 2.19.

Question (15)

. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).

Solution

IN a meeting 70% favors and 30% oppose a proposal.
X = 0 for oppose and X = 1 for favour.
X P(X) XP(X) X2P(X)
0 0.3 0 0
1 0.7 0.7 0.7
Total 1 0.7 0.7

E(X) = ∑XP(X)
= 0.7
E(X2) = ∑X2P(X)
= 0.7
V(X) = E(X2) -[ E(X)]2
= 0.7 - (0.7)2
= 0.7 - 0.49
= 0.21

Question (16)

The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1 (B) 2 (C) 5 (D) 8

Solution

X = 1, 2, 5.
X P(X) XP(X)
1 3/6 3/6
2 2/6 4/6
5 1/6 5/6
Total 1 12/6

Mean E(X) = ∑XP(X)
=12/6
= 2
So B is the correct answer.

Question (17)

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
(A) 37/221 (B) 5/13 (C) 1/13 (D) 2/13

Solution

Two cards are drawn from a pack of cards.
the probability of ace card = 4/52 = 1/13
X ; the number of ace card
X = 0, 1, 2.
X P(X) XP(X)
0 144/169 0
1 24/169 24/169
2 1/169 2/169
Total 1 26/169

E(X) = ∑XP(X)
= 26/169
= 2/13
D is the correct answer.
Exercise 13.3 ⇐
⇒ Exercise 13.5