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NCERT solution class 12 chapter 7 Probability exercise 7.5 mathematics part 2

EXERCISE 7.5


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, 

X has a binomial distribution.

Therefore, P (X = x) = 

(i) P (5 successes) = P (X = 5)

(ii) P(at least 5 successes) = P(X ≥ 5)

(iii) P (at most 5 successes) = P(X ≤ 5)

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

Clearly, X has the binomial distribution with n = 4, 

∴ P (2 successes) = P (X = 2)

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.

X has a binomial distribution with n = 10 and

P(X = x) =

P (not more than 1 defective item) = P (X ≤ 1)

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.

X has a binomial distribution with n = 5 and

(i) P (all five cards are spades) = P(X = 5)

(ii) P (only 3 cards are spades) = P(X = 3)

(iii) P (none is a spade) = P(X = 0)

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

X has a binomial distribution with n = 5 and p = 0.05

(i) P (none) = P(X = 0)

(ii) P (not more than one) = P(X ≤ 1)

(iii) P (more than 1) = P(X > 1)

(iv) P (at least one) = P(X ≥ 1)

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

P (none marked with 0) = P (X = 0)

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 = 

X has a binomial distribution with = 20 and p = 

P (at least 12 questions answered correctly) = P(X ≥ 12)

Question 8:

Suppose X has a binomial distribution. 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 is.

Therefore, n = 6 and

It can be seen that P(X = x) will be maximum, if will be maximum.

The value ofis 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, p

Clearly, X has a binomial distribution with n = 5 and p

P (guessing more than 4 correct answers) = P(X ≥ 4)

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?

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

(a) P (winning at least once) = P (X ≥ 1)

(b) P (winning exactly once) = P(X = 1)

(c) P (at least twice) = P(X ≥ 2)

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, 

Clearly, X has the probability distribution with n = 7 and 

P (getting 5 exactly twice) = P(X = 2)

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

Clearly, X has a binomial distribution with n = 6

P (at most 2 sixes) = P(X ≤ 2)

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% =

P (selecting 9 defective articles) =

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) 

(C) 

(D) 

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

Clearly, X has a binomial distribution with n = 5 and 

P (none of the bulbs is defective) = P(X = 0)

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

(A)  (B) 

(C)  (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, q

Clearly, X has a binomial distribution with n = 5 and 

P (four students are swimmers) = P(X = 4) 

Therefore, the correct answer is A.


 

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