NCERT Solutions for Class 12 Maths Chapter 13 – Probability Ex 13.5
Page No 576:
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?
Answer:
The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.
Probability of getting an odd number in a single throw of a die is, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m254b8ad7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m254b8ad7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_2162836d.gif)
X has a binomial distribution.
Therefore, P (X = x) = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1acdbc14.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1acdbc14.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m13aafad.gif)
(i) P (5 successes) = P (X = 5)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6b51ac2f.gif)
(ii) P(at least 5 successes) = P(X ≥ 5)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_e4158c0.gif)
(iii) P (at most 5 successes) = P(X ≤ 5)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7837/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3827d35f.gif)
Page No 577:
Question 2:
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Answer:
The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.
Probability of getting doublets in a single throw of the pair of dice is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1b38afcb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m62b6a5b3.gif)
Clearly, X has the binomial distribution with n = 4, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m51c87464.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m51c87464.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_5cbfebfe.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m3f5adf61.gif)
∴ P (2 successes) = P (X = 2)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7839/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_55be52d0.gif)
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?
Answer:
Let X denote the number of defective items in a sample of 10 items drawn successively. Since the drawing is done with replacement, the trials are Bernoulli trials.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6813b8a1.gif)
X has a binomial distribution with n = 10 and![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_6dc5e870.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_6dc5e870.gif)
P(X = x) =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m73e8d9ef.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m73e8d9ef.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m311075e5.gif)
P (not more than 1 defective item) = P (X ≤ 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7841/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m4e1e812f.gif)
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?
Answer:
Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.
In a well shuffled deck of 52 cards, there are 13 spade cards.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6416bc.gif)
X has a binomial distribution with n = 5 and![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3f13a6d0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3f13a6d0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_11442331.gif)
(i) P (all five cards are spades) = P(X = 5)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m81782de.gif)
(ii) P (only 3 cards are spades) = P(X = 3)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_2a4f820.gif)
(iii) P (none is a spade) = P(X = 0)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7843/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m52de3081.gif)
Question 5:
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is 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.
Answer:
Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.
It is given that, p = 0.05
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7845/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7554b486.gif)
X has a binomial distribution with n = 5 and p = 0.05
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7845/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m604728cd.gif)
(i) P (none) = P(X = 0)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7845/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m34037b3b.gif)
(ii) P (not more than one) = P(X ≤ 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7845/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m5fc18682.gif)
(iii) P (more than 1) = P(X > 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7845/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m5f061e7f.gif)
(iv) P (at least one) = P(X ≥ 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7845/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_74bf7b93.gif)
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?
Answer:
Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials.
X has a binomial distribution with n = 4 and![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7848/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_46738c69.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7848/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_46738c69.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7848/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1cf6cb3f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7848/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_40fd4879.gif)
P (none marked with 0) = P (X = 0)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7848/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_78bdf320.gif)
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.
Answer:
Let X represent the number of correctly answered questions out of 20 questions.
The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.
∴ p = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m7b8ceffd.gif)
X has a binomial distribution with n = 20 and p = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m3947d8ba.gif)
P (at least 12 questions answered correctly) = P(X ≥ 12)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7849/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m413c7709.gif)
Question 8:
Suppose X has a binomial distribution
. Show that X = 3 is the most likely outcome.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_6e1f6054.gif)
(Hint: P(X = 3) is the maximum among all P (xi), xi = 0, 1, 2, 3, 4, 5, 6)
Answer:
X is the random variable whose binomial distribution is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_518c59d.gif)
Therefore, n = 6 and![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m2bf18ad.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m2bf18ad.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m69d8a036.gif)
It can be seen that P(X = x) will be maximum, if
will be maximum.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3f9d4031.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m5c29518b.gif)
The value of
is maximum. Therefore, for x = 3, P(X = x) is maximum.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7852/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4da805f1.gif)
Thus, X = 3 is the 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?
Answer:
The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.
Probability of getting a correct answer is, p![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1da506e9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1da506e9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m51531a81.gif)
Clearly, X has a binomial distribution with n = 5 and p![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1da506e9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1da506e9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_35688ca9.gif)
P (guessing more than 4 correct answers) = P(X ≥ 4)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7854/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4efbf40c.gif)
Question 10:
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is
. What is the probability that he will in a prize (a) at least once (b) exactly once (c) at least twice?
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1415a943.gif)
Answer:
Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.
Clearly, X has a binomial distribution with n = 50 and![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_5053b775.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_5053b775.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m537391d6.gif)
(a) P (winning at least once) = P (X ≥ 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m37a216f4.gif)
(b) P (winning exactly once) = P(X = 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3a891296.gif)
(c) P (at least twice) = P(X ≥ 2)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7856/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_42c14c21.gif)
Page No 578:
Question 11:
Find the probability of getting 5 exactly twice in 7 throws of a die.
Answer:
The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.
Probability of getting 5 in a single throw of the die, p ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_68437504.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_68437504.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m62b6a5b3.gif)
Clearly, X has the probability distribution with n = 7 and p ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_68437504.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_68437504.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m51a0d16d.gif)
P (getting 5 exactly twice) = P(X = 2)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7858/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m17a929a2.gif)
Question 12:
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Answer:
The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.
Probability of getting six in a single throw of die, p![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7860/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_68437504.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7860/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_68437504.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7860/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m62b6a5b3.gif)
Clearly, X has a binomial distribution with n = 6
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7860/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7168953e.gif)
P (at most 2 sixes) = P(X ≤ 2)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7860/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3c8d4544.gif)
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?
Answer:
The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.
Clearly, X has a binomial distribution with n = 12 and p = 10% =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m958c3e4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m958c3e4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1cf6cb3f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m59575170.gif)
P (selecting 9 defective articles) =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m61abdcf2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m61abdcf2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7862/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_540e343d.gif)
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) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1e93c9db.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1e93c9db.gif)
(C) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m34a2d66.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m34a2d66.gif)
(D) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6d78cbca.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6d78cbca.gif)
Answer:
The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.
Probability of getting a defective bulb, p![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m64b86949.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m64b86949.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1cf6cb3f.gif)
Clearly, X has a binomial distribution with n = 5 and ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_46738c69.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_46738c69.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_41b26fc.gif)
P (none of the bulbs is defective) = P(X = 0)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7864/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_77e7a812.gif)
The correct answer is C.
Question 15:
The probability that a student is not a swimmer is
. Then the probability that out of five students, four are swimmers is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3d0b22e5.gif)
(A)
(B) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m3b8f6c9b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1183573d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m3b8f6c9b.gif)
(C)
(D) None of these
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_18951da1.gif)
Answer:
The repeated selection of students who are swimmers are Bernoulli trials. Let X denote the number of students, out of 5 students, who are swimmers.
Probability of students who are not swimmers, q![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m55a4b08b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m55a4b08b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m587bf2ca.gif)
Clearly, X has a binomial distribution with n = 5 and ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_14a9695e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_14a9695e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_2d0783ac.gif)
P (four students are swimmers) = P(X = 4) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_11c5cbd6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7866/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_11c5cbd6.gif)
Therefore, the correct answer is A.