12th NCERT Probability Exercise 13.5 Questions 15
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Question (1)

A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?

Solution

' getting an odd number ' is success.
p = probability of success. = 3/6 = 1/2 , q = 1 - p = 1/2, n = 6.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ (i) 5 success = P(X = 5 )
$= 6{C_5}{\left( {\frac{1}{2}} \right)^5}\left( {\frac{1}{2}} \right)$ $= 6 \times \frac{1}{{32}} \times \frac{1}{2}$ $= \frac{3}{{32}}$ (ii) at least 5 success = P(X ≥ 5)
= P(X=5) + P(X=6)
$= 6{C_5}{\left( {\frac{1}{2}} \right)^5}{\left( {\frac{1}{2}} \right)^1} + 6{C_6}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^0}$ $= 6 \times \frac{1}{{32}} \times \frac{1}{2} + 1 \times \frac{1}{{64}} \times 1$ $= \frac{6}{{64}} + \frac{1}{{64}}$ $= \frac{7}{{64}}$ (iii) at most 5 success = P(X ≤5)
= 1 - P(6)
$= 1 - 6{C_6}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^0}$ $= 1 - 1 \times \frac{1}{{64}} \times 1$ $= 1 - \frac{1}{{64}}$ $= \frac{{63}}{{64}}$

Question (2)

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.

Solution

A pair of dice are thrown 4 times. n = 4
Getting a doublet is success,
p = probability of success = 6/36 = 1/6 , q = 1 - p = 5/6.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P(X=2)= $= 4{C_2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^2}$ $= 6 \times \frac{1}{{36}} \times \frac{{25}}{{36}}$ $= \frac{{25}}{{216}}$

Question (3)

There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Solution

There are 5% defective items in bulk of items.
p = probability of success = 0.05 = 1/20 , q = 1 - p = 19/20, n = 10
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P( not more than 2 defective items)
=P(X<2)
= P(X=0) + P(X=1)
$= 10{C_0}{\left( {\frac{1}{{20}}} \right)^0}{\left( {\frac{{19}}{{20}}} \right)^{10}} + 10{C_1}{\left( {\frac{1}{{20}}} \right)^1}{\left( {\frac{{19}}{{20}}} \right)^9}$ $= 1 \times 1 \times {\left( {\frac{{19}}{{20}}} \right)^{10}} + 10 \times \frac{1}{{20}} \times {\left( {\frac{{19}}{{20}}} \right)^9}$ $= {\left( {\frac{{19}}{{20}}} \right)^9}\left[ {\frac{{19}}{{20}} + \frac{{10}}{{20}}} \right]$ ${\left( {\frac{{19}}{{20}}} \right)^9}\left( {\frac{{29}}{{20}}} \right)$ $= \left( {\frac{{29}}{{20}}} \right){\left( {\frac{{19}}{{20}}} \right)^9}$

Question (4)

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that (i) all the five cards are spades? (ii) only 3 cards are spades? (iii) none is a spade?

Solution

Five cards are drawn from the pack of cards. n = 5
p = probability of sucess = getting a spade card.
p = 13/52 = 1/4, q = 1 - p = 3/4
By Binomial expansion, $P(X = x) = n{C_x}{p^x}{q^{n - x}}$ (i) all 5 cards are of spade.
P(X = 5) $= 5{C_5}{\left( {\frac{1}{4}} \right)^5}{\left( {\frac{3}{4}} \right)^0}$ $= 1 \times \frac{1}{{1024}} \times 1$ $= \frac{1}{{1024}}$ (ii) only 3 cards are of spade.
P(X=3) $= 5{C_3}{\left( {\frac{1}{4}} \right)^3}{\left( {\frac{3}{4}} \right)^2}$ $= 10 \times \frac{1}{{64}} \times \frac{9}{{16}}$ $= \frac{{45}}{{512}}$ (iii) none is spade card
P(X=0) $= 5{C_0}{\left( {\frac{1}{4}} \right)^0}{\left( {\frac{3}{4}} \right)^5}$ $= 1 \times 1 \times \frac{{243}}{{1024}}$ $= \frac{{243}}{{1024}}$

Question (5)

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use.

Solution

p = probability of success = a bulb produced by a factory will fuse after 150 days of use
p = 0.05, q = 1 - p = 0.95, n = 5.
By Binomial expansion, $P(X = x) = n{C_x}{p^x}{q^{n - x}}$ (i) none
P(X=0) $= 5{C_0}{\left( {0.05} \right)^0}{\left( {0.95} \right)^5}$ $= 1 \times 1 \times {\left( {0.95} \right)^5}$ $= {\left( {0.95} \right)^5}$ (ii) Not more than 1
= P(X ≤1)
= P(X=0)+P(X=1)
$= \left[ {5{C_0}{{\left( {0.05} \right)}^0}{{\left( {0.95} \right)}^5} + 5{C_1}{{\left( {0.05} \right)}^1}{{\left( {0.95} \right)}^4}} \right]$ $= \left[ {{{\left( {0.95} \right)}^5} + 5\left( {0.05} \right){{\left( {0.95} \right)}^4}} \right]$ $= {\left( {0.95} \right)^4}\left( {0.95 + 0.25} \right)$ $= {\left( {0.95} \right)^4}\left( {1.2} \right)$ (iii) more than 1
P(X >1)
= P(X=2)+P(X=3)+P(X=4)+P(X=5)
= 1 -[ P(X=0)+P(X=1)] $= 1 - \left[ {5{C_0}{{\left( {0.05} \right)}^0}{{\left( {0.95} \right)}^5} + 5{C_1}{{\left( {0.05} \right)}^1}{{\left( {0.95} \right)}^4}} \right]$ $= 1 - \left[ {{{\left( {0.95} \right)}^5} + 5\left( {0.05} \right){{\left( {0.95} \right)}^4}} \right]$ $= 1 - {\left( {0.95} \right)^4}\left( {0.95 + 0.25} \right)$ $= 1 - {\left( {0.95} \right)^4}\left( {1.2} \right)$ (iv) At least 1
= P(X ≥1)
= 1 - P(X=0) $= 1 - \left[ {5{C_0}{{\left( {0.05} \right)}^0}{{\left( {0.95} \right)}^5}} \right]$ $= 1 - {\left( {0.95} \right)^5}$

Question (6)

A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Solution

A ball marked with digis 0 to 9. , n = 4
p = probability of success = not getting a ball with digit 0
∴ p = 1/10, q = 1 - p = 9/10
By Binomial expansion, $P(X = x) = n{C_x}{p^x}{q^{n - x}}$ None is marked with 0
P(X = 0) $= \left[ {4{C_0}{{\left( {\frac{1}{{10}}} \right)}^0}{{\left( {\frac{9}{{10}}} \right)}^4}} \right]$ $= 1 \times 1 \times {\left( {\frac{9}{{10}}} \right)^4}$ $= {\left( {\frac{9}{{10}}} \right)^4}$

Question (7)

In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Solution

p =probability of success
= giving the correct answer to question
p = 1/2, q = 1 - p = 1/2, n = 20
At least 12 question correctly
= P(X ≥12)
= P(X=12)+P(X=13)+P(X=14) +......+P(X=20)
By Binomial expansion, $P(X = x) = n{C_x}{p^x}{q^{n - x}}$ $= 20{C_{12}}{\left( {\frac{1}{2}} \right)^{12}}{\left( {\frac{1}{2}} \right)^8} + 20{C_{13}}{\left( {\frac{1}{2}} \right)^{13}}{\left( {\frac{1}{2}} \right)^7} + 20{C_{14}}{\left( {\frac{1}{2}} \right)^{14}}{\left( {\frac{1}{2}} \right)^6} + ............ + 20{C_{20}}{\left( {\frac{1}{2}} \right)^{20}}{\left( {\frac{1}{2}} \right)^0}$ $= {\left( {\frac{1}{2}} \right)^{20}}\left[ {20{C_{12}} + 20{C_{13}} + 20{C_{14}} + ....... + 20{C_{20}}} \right]$

Question (8)

Suppose X has a binomial distribution B (6,1/2). Show that X = 3 is the most likely outcome. (Hint: P(X = 3) is the maximum among all P(xi), xi = 0, 1, 2, 3, 4, 5, 6)

Solution

X is the binomial with n = 6, p = 1/2, q = 1 - p = 1/2.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P(X= 0) $= 6{C_0}{\left( {\frac{1}{2}} \right)^0}{\left( {\frac{1}{2}} \right)^6}$ $= {\left( {\frac{1}{2}} \right)^6}$ $= \frac{1}{{64}}$ P(X=1) $= 6{C_1}{\left( {\frac{1}{2}} \right)^1}{\left( {\frac{1}{2}} \right)^5}$ $= 6 \times {\left( {\frac{1}{2}} \right)^6}$ $= \frac{6}{{64}}$ P(X = 2)$= 6{C_2}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^4}$ $= 15 \times {\left( {\frac{1}{2}} \right)^6}$ $= \frac{{15}}{{64}}$ P(X = 3)$= 6{C_3}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^3}$ $= 20 \times {\left( {\frac{1}{2}} \right)^6}$ $= \frac{{20}}{{64}}$ P(X = 4)$= 6{C_4}{\left( {\frac{1}{2}} \right)^4}{\left( {\frac{1}{2}} \right)^2}$ $= 15 \times {\left( {\frac{1}{2}} \right)^6}$ $= \frac{{15}}{{64}}$ P(X = 5)$= 6{C_5}{\left( {\frac{1}{2}} \right)^5}{\left( {\frac{1}{2}} \right)^1}$ $= 6 \times {\left( {\frac{1}{2}} \right)^6}$ $= \frac{{6}}{{64}}$ P(X = 6)$= 6{C_6}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^0}$ $= 1 \times {\left( {\frac{1}{2}} \right)^6}$ $= \frac{{1}}{{64}}$ Since P(X = 3) have maximum value of probability, X = 3 is most likely outcome.

Question (9)

On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Solution

p = probability of success
= giving correct answer by guessing = 1/3
p = 1/3, q = 1 - p = 2/3, n = 5.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P( 4 or more correct ansers)
= P(X ≥4)
= P(X = 4) + P(X = 5)
$= 5{C_4}{\left( {\frac{1}{3}} \right)^4}{\left( {\frac{2}{3}} \right)^1} + 5{C_5}{\left( {\frac{1}{3}} \right)^5}{\left( {\frac{2}{3}} \right)^0}$ $= 5 \times \frac{1}{{81}} \times \frac{2}{3} + 1 \times \frac{1}{{243}} \times 1$ $= \frac{{10 + 1}}{{243}}$ $= \frac{{11}}{{243}}$

Question (10)

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100. What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice?

Solution

n = 50, p = probability of winning = 1/100, q = 1 - p = 99/100
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ (a) at least once.
P( X ≥ 1)
= 1 - P(X = 0)
$= 1 - 50{C_0}{\left( {\frac{1}{{100}}} \right)^0}{\left( {\frac{{99}}{{100}}} \right)^{50}}$ $= 1 - {\left( {\frac{{99}}{{100}}} \right)^{50}}$ (b) Exactly once.
P(X = 0 )
$= 50{C_1}{\left( {\frac{1}{{100}}} \right)^1}{\left( {\frac{{99}}{{100}}} \right)^{49}}$ $= 50 \times \frac{1}{{100}} \times {\left( {\frac{{99}}{{100}}} \right)^{49}}$ $= \frac{1}{2}{\left( {\frac{{99}}{{100}}} \right)^{49}}$ (c) at least twice
P( X ≥ 2)
= 1 - [ P(X=0) + P(X=1)]
$= 1 - \left[ {50{C_0}{{\left( {\frac{1}{{100}}} \right)}^0}{{\left( {\frac{{99}}{{100}}} \right)}^{50}} + 50{C_1}{{\left( {\frac{1}{{100}}} \right)}^1}{{\left( {\frac{{99}}{{100}}} \right)}^{49}}} \right]$ $= 1 - \left[ {1 \times 1 \times {{\left( {\frac{{99}}{{100}}} \right)}^{50}} + 50 \times \frac{1}{{100}} \times {{\left( {\frac{{99}}{{100}}} \right)}^{49}}} \right]$ $= 1 - {\left( {\frac{{99}}{{100}}} \right)^{49}}\left[ {\frac{{99}}{{100}} + \frac{{50}}{{100}}} \right]$ $= 1 - \frac{{149}}{{100}}{\left( {\frac{{99}}{{100}}} \right)^{49}}$

Question (11)

. Find the probability of getting 5 exactly twice in 7 throws of a die.

Solution

A die is thrown 7 times. n = 7.
getting 5 is success, p = 1/6, q = 1 - p = 5/6.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P( exactly 2)= P(X = 2) $= 7{C_2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^5}$ $= 21 \times \frac{1}{{36}} \times {\left( {\frac{5}{6}} \right)^5}$ $= \frac{7}{{12}}{\left( {\frac{5}{6}} \right)^5}$

Question (12)

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Solution

A die is thrown 6 times, n = 6.
getting 6 is a success, p = 1/6, q = 1 - p = 5/6.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P(At most 2 sixes.) = P(X ≤2)
P(X = 0)+P(X = 1) + P(X = 2) $= 6{C_0}{\left( {\frac{1}{6}} \right)^0}{\left( {\frac{5}{6}} \right)^6} + 6{C_1}{\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^5} + 6{C_2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^4}$ $= {\left( {\frac{5}{6}} \right)^4}\left[ {\frac{{25}}{{36}} + \frac{{30}}{{36}} + \frac{{15}}{{36}}} \right]$ $= {\left( {\frac{5}{6}} \right)^4}\left( {\frac{{70}}{{36}}} \right)$ $= \frac{{35}}{{18}}{\left( {\frac{5}{6}} \right)^4}$

Question (13)

. It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

Solution

getting a defective is success, p = 1/10, q = 1 - p = 9/10, n = 12.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P( X = 9 ) $= 12{C_9}{\left( {\frac{1}{{10}}} \right)^9}{\left( {\frac{9}{{10}}} \right)^3}$ $= \frac{{12 \times 11 \times 10}}{{3 \times 2}} \times \frac{1}{{{{10}^9}}} \times \frac{{{9^3}}}{{{{10}^3}}}$ $= \frac{{220 \times {9^3}}}{{{{10}^{12}}}}$ $= \frac{{22 \times {9^3}}}{{{{10}^{11}}}}$

Question (14)

. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(A) 10−1 (B) (1/2)5 (C) (9/10)5 (D) 9/10

Solution

p = probability of defective bulb
p = 1/10, q = 1 - p = 9/10, n = 5.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P( X = 0) $= 5{C_0}{\left( {\frac{1}{{10}}} \right)^0}{\left( {\frac{9}{{10}}} \right)^5}$ $= 1 \times 1 \times {\left( {\frac{9}{{10}}} \right)^5}$ $= {\left( {\frac{9}{{10}}} \right)^5}$ So C is the correct answer.

Question (15)

The probability that a student is not a swimmer is 1/5. Then the probability that out of five students, four are swimmers is
(A)$5{C_4}{\left( {\frac{4}{5}} \right)^4}{\left( {\frac{1}{5}} \right)^1}$ (B) ${\left( {\frac{4}{5}} \right)^4}{\left( {\frac{1}{5}} \right)^1}$ (C) $5{C_1}{\left( {\frac{4}{5}} \right)^4}{\left( {\frac{1}{5}} \right)^1}$ (D) None of these

Solution

let p = probability that student is swimmer .
probability that he is not swimmer = 1/5, so p = 4/5, q = 1 - p = 1/5, n = 5.
$P(X = x) = n{C_x}{p^x}{q^{n - x}}$ P( X = 4) $= 5{C_4}{\left( {\frac{4}{5}} \right)^4}{\left( {\frac{1}{5}} \right)^1}$ $= 5{C_1}{\left( {\frac{4}{5}} \right)^4}{\left( {\frac{1}{5}} \right)^1}$ So A and C are the correct answers.