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Exercise 13.5 - Chapter 13 Probability class 12 ncert solutions Maths - SaraNextGen [2024-2025]


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

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_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_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

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_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 andhttps://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_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 andhttps://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 andhttps://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_m7b8ceffd.gif

X has a binomial distribution with = 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_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 distributionhttps://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. 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)

Answer:

X is the random variable whose binomial distribution ishttps://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 andhttps://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 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.gifwill 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_m5c29518b.gif

The value ofhttps://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.gifis maximum. Therefore, for x = 3, P(X = x) is maximum.

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, phttps://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 phttps://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 ishttps://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. What is the probability that he will in a prize (a) at least once (b) exactly once (c) at least twice?

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 andhttps://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

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, 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 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, phttps://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 = 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_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_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

(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

(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

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, phttps://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_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 ishttps://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. Then the probability that out of five students, four are swimmers is

(A) 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 (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

(C) 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 (D) None of these

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, qhttps://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_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

Therefore, the correct Answer is A.

Also Read : Miscellaneous-Exercise-Chapter-13-Probability-class-12-ncert-solutions-Maths

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