**Chapter 8 Binomial Distribution Ex 8.1**

## Chapter 8 Binomial Distribution Ex 8.1

**Question 1.A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of(i) 5 successes(ii) at least 5 successes(iii) at most 5 successes.Solution:**

Let X = number of successes, i.e. number of odd numbers.

p = probability of getting an odd number in a single throw of a die

The p.m.f. of X is given by

Hence, the probability of 5 successes is 3/32.

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

= p(5) + p(6)

Hence, the probability of at least 5 successes is 7/64.

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

= 1 – P[X > 5]

Hence, the probability of at most 5 successes is 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:**

Let X = number of doublets.

p = probability of getting a doublet when a pair of dice is thrown

The p.m.f. of X is given by

Hence, the probability of two successes is 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:**

Let X = number of defective items.

p = probability of defective item

The p.m.f. of X is given by

P(sample of 10 items will include not more than one defective item) = P[X ≤ 1]

Hence, the probability that a sample of 10 items will include not more than one

**Question 4.Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that(i) all the five cards are spades(ii) only 3 cards are spades(iii) none is a spade.Solution:**

Let X = number of spade cards.

p = probability of drawing a spade card from a pack of 52 cards.

Since there are 13 spade cards in the pack of 52 cards.

The p.m.f. of X is given by

(i) P(all five cards are spade)

Hence, the probability of all the five cards are spades =1-1024

(ii) P(only 3 cards are spade) = P[X = 3]

Hence, the probability of only 3 cards are spades = 45/512

(iii) P(none of cards is spade) = P[X = 0]

Hence, the probability of none of the cards is a spade = 243/1024

**Question 5.The probability of 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:**

Let X = number of fuse bulbs.

p = probability of a bulb produced by a factory will fuse after 150 days of use.

∴ p = 0.05

and q = 1 – p = 1 – 0.05 = 0.95

Given: n = 5

∴ X ~ B(5, 0.05)

The p.m.f. of X is given by

(i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0]

= p(0)

Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)

^{5}.

(ii) P(not more than one bulb will fuse after 150 days of j use) = P[X ≤ 1]

= p(0) + p(1)

(iii) P(more than one bulb fuse after 150 days)

= P[X > 1]

= 1 – P[X ≤ 1]

= 1 – (1.2)(0.95

Hence, the probability that more than one bulb fuse after 150 days = 1 – (1.2)(0.95.

(iv) P(at least one bulb fuse after 150 days)

= P[X ≥ 1]

= 1 – P[X = 0]

= 1 – p(0)

**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:**

Let X = number of balls marked with digit 0.

p = probability of drawing a ball from 10 balls marked with the digit 0.

The p.m.f. of X is given by

P(none of the ball marked with digit 0) = P(X = 0)

Hence, the probability that none of the bulb marked with digit 0 is

**Question 7.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:**

Let X = number of correct answers.

p = probability that a candidate gets a correct answer from three possible answers.

The p.m.f. of X is given by

P(four or more correct answers) = P[X ≥ 4]

= p(4) + p(5)

Hence, the probability of getting four or more correct answers = 11/243.

**Question 8.A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100, find the probability that he will win a prize(i) at least once(ii) exactly once(iii) at least twice.Solution:**

Let X = number of winning prizes.

p = probability of winning a prize

(i) P(a person wins a prize at least once)

Hence, probability of winning a prize at least once = 1 –

(ii) P(a person wins exactly one prize) = P[X = 1] = p(1)

(iii) P(a persons wins the prize at least twice) = P[X ≥ 2]

= 1 – P[X < 2]

= 1 – [p(0) + p(1)]

**Question 9.In a box of floppy discs, it is known that 95% will work. A sample of three of the discs is selected at random. Find the probability that (i) none (ii) 1 (iii) 2 (iv) all 3 of the sample will work.Solution:**

Let X = number of working discs.

p = probability that a floppy disc works

The p.m.f. of X is given by

(i) P(none of the floppy discs work) = P(X = 0)

(ii) P(exactly one floppy disc works) = P(X = 1)

(iii) P(exactly two floppy discs work) = P(X = 2)

(iv) P(all 3 floppy discs work) = P(X = 3)

Hence, the probability that all 3 floppy discs work = .

**Question 10.Find the probability of throwing at most 2 sixes in 6 throws of a single die.Solution:**

Let X = number of sixes.

p = probability that a die shows six in a single throw

The p.m.f. of X is given by

**Question 11.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:**

Let X = number of defective articles.

p = probability of defective articles.

The p.m.f. of X is given by

**Question 12.Given X ~ B(n, P)(i) If n = 10 and p = 0.4, find E(x) and Var(X).(ii) If p = 0.6 and E(X) = 6, find n and Var(X).(iii) If n = 25, E(X) = 10, find p and SD(X).(iv) If n = 10, E(X) = 8, find Var(X).Solution:**

(i) Given: n = 10 and p = 0.4

∴ q = 1 – p = 1 – 0.4 = 0.6

∴ E(X) = np = 10(0.4) = 4

Var(X) = npq = 10(0.4)(0.6) = 2.4

Hence, E(X) = 4, Var(X) = 2.4.

(ii) Given: p = 0.6, E (X) = 6

E(X) = np

6 = n(0.6)

n = 6/0.6 = 10

Now, q = 1 – p = 1 – 0.6 = 0.4

∴ Var(X) = npq = 10(0.6)(0.4) = 2.4

Hence, n = 10 and Var(X) = 2.4.

(iii) Given: n = 25, E(X) = 10

E(X) = np

10 = 25p

∴ SD(X) = √Var(X) = √6

Hence, p = 25 and S.D.(X) = √6.

(iv) Given: n = 10, E(X) = 8

E(X) = np

8 = 10p