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Class 12 RD Sharma Solutions – Chapter 33 Binomial Distribution – Exercise 33.1 | Set 1

  • Last Updated : 21 Jul, 2021

Question 1. There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.

Solution:

Let us consider X be the number of defective items in a sample of 8 items. 

So, a binomial distribution follows by x with n = 8.

and Probability of getting a defective item(p) = 0.06 

The probability of getting a defective item (1 – p) = 0.94



So, P(X = r) = 8Cr (0.06)r (0 . 94 )8 – r, where r = 0, 1, 2, 3, . . . 8

So, the required probability is

= P (X < 1) 

= P(X = 0) + P(X = 1)

= 8C0 (0.06)0 (0.94)8-0 + 8C1 (0.06)1 (0.94 )8-1

= (0.94)8 + 8 (0.06) (0.94)

= (0.94)7 {0.94 + 0.48}

= 1.42 (0.94)7



Question 2. A coin is tossed 5 times. What is the probability of getting at least 3 heads?

Solution:

Let us consider X be the number of heads in 5 tosses.

So, a binomial distribution follows by x with n = 5

Probability of getting a head(p) = 1/2

And q = 1 – p = 1 – 1/2 = 1/2

P(X = r) = 5Cr (1/2)r (1/2)5-r , where r = 0, 1, 2 . . . 5

So, the required probability is

= P(X > 3)

= P(X = 3) + P(X = 4) + P(X = 5)

= 5C3 (1/2)3 (1/2)5-3 + 5C4 (1/2)1 (1/2)5-1 + 5C5 (1/2)5 (1/2)5-0

= 10 (1/2)5 + 5 (1/2)5 + 1 (1/2)5

= (1/2)5 (10 + 5 + 1)

= (1/2)5 (16)

= 1/2

Question 3. A coin is tossed 5 times. What is the probability that tail appears an odd number of times?

Solution:

Let us consider X be the number of tails when a coin is tossed 5 times.

So, a binomial distribution follows by x with n  = 5

Probability of getting head(p) = 1/2.

Also, q = 1 – p = 1/2

P(X = r) = 5C3 (1/2)r (1/2)n-r = 5Cr (1/2)5



So, the required probability is

= P(X = 1) + P(X = 3) + P(X = 5)

= 5C1 (1/2)5 + 5C3 (1/2)5 + 5C5 (1/2)5

= (1/2)5 [5 + 10 + 1]

= 16/32

= 1/2

Question 4. A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?

Solution:

Let us consider X be the number of successes in 6 throws of the two dice.

So the probability of success is equal to the probability of getting a total of 9  

= Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) from 36 outcomes

Here, p = 4/36 = 1/9

Also q = 1 – p = 8/9 and n = 6

Now, a binomial distribution X follows with n = 6, p = 1/9, and q = 8/9

P(X = r) = 6Cr (1/9)r (8/9)6-r

The required probability is

= P(X > 5)

= P(X = 5) + P(X = 6)

^{6}{}{C}_5 \left( \frac{1}{9} \right)^5 \left( \frac{8}{9} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{9} \right)^6 \left( \frac{8}{9} \right)^{6 - 6}

\frac{6(8) + 1}{9^6}

= 49/96



Question 5. A fair coin is tossed 8 times, find the probability: 

(i) of exactly 5 heads.

Solution:

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n  = 8.

Here, p = 1/2 and q = 1 – 1/2 = 1/2

So, P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

= 8Cr (1/2)8 , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is 

P(X = 5) = 8C5 (1/2)8

= 56/256

= 7/32

(ii) of at least six heads.

Solution:

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1 – 1/2 = 1/2

So, P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

= 8Cr (1/2)8 , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is 

P(X = 5) = 8C5 (1/2)8

= 56/256

= 7/32



(iii) of at most six heads.

Solution:

Let us consider X denotes the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here p = 1/2 and q = 1 – 1/2 = 1/2

P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

= 8Cr (1/2)8 , r = 0, 1, 2, . . . , 8

so the probability of obtaining at most 6 heads is

P (X < 6) = 1 – [P(X = 7) + P(X = 8)]

1 - \left[ {}^8 C_7 \left( \frac{1}{2} \right)^8 +^8 C_8 \left( \frac{1}{2} \right)^8 \right]

= 1 – (8/256 + 1/256)

= 1 – 9/256

= 247/256

Question 6. Find the probability of 4 turning up at least once in two tosses of a fair die.

Solution:

Let us consider X denotes the probability of getting 4 in two tosses of a fair die.

Now, a binomial distribution X follows with n = 2.

Here, p = 1/6 and q = 5/6

P(X = r) = ^{2}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{2 - r}

So, the probability of obtaining 4 at least once is

P(X > 1) = 1 – P(X = 0)

= 1 – ^{2}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{2 - 0}



= 1 – 25/36

= 11/36

Question 7. A coin is tossed 5 times. What is the probability that head appears an even number of times?

Solution:

Let us consider X denotes the number of heads that appear when a coin is tossed 5 times.

Now, a binomial distribution X follows with n = 5 

Here p = q = 1/2.

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r}

= 5Cr (1/2)5

P (head appears an even number of  times) = P(X = 0) + P(X = 2) + P(X = 4)

^{5}{}{C}_0 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_2 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_4 \left( \frac{1}{2} \right)^5

\frac{1 + 10 + 5}{2^5}

= 16/32

= 1/2

Question 8. The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?

Solution:

Let us consider X denotes the number of times the target is hit. 

Now, a binomial distribution X follows with n = 7.

Here, p = 1/4 and q = 3/4.

P(X = r) = ^{7}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{7 - r}

P(hit the target at least twice) = P(X > 2) 

= 1 – {P(X = 0) + P(X = 1)}



1 -^{7}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{7 - 0} - ^{7}{}{C}_1 \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^{7 - 1}

= 1 – (3/4)7 – 7 (1/4) (3/4)6

= 1 – 1/16384(2187 + 5103)

= 1 – 3645/8192

= 4547/8192

Question 9. Assume that on average one telephone number out of 15 called between 2 P.M. and 3 P.M. on weekdays is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?

Solution:

Let us consider X denotes the number of busy calls for 6 randomly selected telephone numbers.

Now, a binomial distribution X follows with n = 6.

Here, p = one out of 15 = 1/15

And q = 1 – 1/15 = 14/15

P(X = r) = ^{6}{}{C}_r \left( \frac{1}{15} \right)^r \left( \frac{14}{15} \right)^{6 - r}

So, the probability that at least 3 of them are busy is 

P(X > 3) = 1 – {P(X = 0) + P(X = 1) + P(X = 2)}

1 - \left\{ ^{6}{}{C}_0 \left( \frac{1}{15} \right)^0 \left( \frac{14}{15} \right)^{6 - 0} + ^{6}{}{C}_1 \left( \frac{1}{15} \right)^1 \left( \frac{14}{15} \right)^{6 - 1} + ^{6}{}{C}_2 \left( \frac{1}{15} \right)^2 \left( \frac{14}{15} \right)^{6 - 2} \right\}

1 - \left\{ \left( \frac{14}{15} \right)^6 + \frac{6}{15} \left( \frac{14}{15} \right)^5 + \frac{1}{15} \left( \frac{14}{15} \right)^4 \right\}

Question 10. If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find P (X ≥ 4).

Solution:

Let us consider X be the number of successes. That is getting 5 or 6 in a throw of die in 6 throws.

Now, a binomial distribution X follows with n = 6

Here, p = of getting 5 or 6 = 1/6 + 1/6 = 1/3

 And q = 1 – p = 2/3



P(X = r) = ^{6}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{6 - r}

P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6)

^{6}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^{6 - 4} +^{6}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{3} \right)^6 \left( \frac{2}{3} \right)^{6 - 6}

= (1/36) (60 + 12 + 1)

= 73/729

Question 11. Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.

Solution:

Let us consider X denotes the number of heads in tossing 8 coins.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1/2

P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r}

= 8Cr (1/2)8

So, the probability of getting at least 6 heads is 

P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)

^{8}{}{C}_6 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8

= 1/28(28 + 8 + 1)

= 37/256

Question 12. Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards.

(i) What is the probability that all the five cards are spades?

Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

p = 13/52 = 1/4 and q = 1 – p = 3/4



P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

P(All cards are spades) = P(X = 5)

^{5}{}{C}_5 \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0

= 1/1024

(ii) What is the probability that only 3 cards are spades? 

Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5. 

Here, p = 13/52 = 1/4 and q = 1 – p = 3/4

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

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

^{5}{}{C}_3 \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^2

= (1/1024) (90)

= 45/512

(iii) What is the probability that none is a spade?

Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

Here p = 13/52 = 1/4 and q = 1 – p = 3/4.

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

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

^{5}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^5



= 243/1024

Question 13. A bag contains 7 red, 5 white, and 8 black balls. Four balls are drawn one by one with replacement.

(i) What is the probability that none is white?

Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

So, the probability that none is white is

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}

= 81/256

(ii) What is the probability that none is white?

Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

so, the probability that none is white is 

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}

= 81/256

(iii) What is the probability that any two are white?

Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

So, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

So, the probability that any two are white is 

P(X = 2) = ^{4}{}{C}_2 \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^{4 - 2}

= 54/256

= 27/128

Question 14. A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.

Solution:



Let us consider X denotes the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.

Now, a binomial distribution X follows with n = 5.

Here, the probability of getting a ticket bearing number divisible by 10(p) = 10/100

= 1/10

And q = 1 – p = 9/10

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r}

So, the probability that all the tickets bear numbers divisible by 10 is

P(X = 5) = ^{5}{}{C}_5 \left( \frac{1}{10} \right)^5 \left( \frac{9}{10} \right)^{5 - 5}

= (1/10)5 (9/10)0

= (1/10)5

Question 15. A bag contains 10 balls, each marked with one of the digits from 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 us consider X denotes the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

Now, a binomial distribution X follows with n = 4.

So, the probability that a ball randomly drawn bears digit 0(p) = 1/10

And q = 1 – p = 9/10

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{4 - r}

P(none bears the digit 0) = P(X = 0)

^{4}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^{4 - 0}

= (9/10)4

Question 16. In a large bulk of items, 5 percent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?

Solution:

Let us consider X be the number of defective items in a sample of 10 items.

Now, a binomial distribution X follows with n = 10.

And the probability of defective items(p) = 5% = 0.05 

And q = 1 – p = 0.95

P(X = r) = 10Cr (0. 05)r (0.95)10-r

So, the probability(sample of 10 items will include not more than one defective item) is 

P(X < 1) = P(X = 0) + P(X = 1) 

= 10C0 (0.05)0 (0.95 ){0-0 + 10C1 (0.05)1 (0.95)10-1

= (0.95)9 (0.95 + 0.5)

= 1.45 (0.95)9

Question 17. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. 

(i) Find the probability that out of 5 such bulbs none will fuse after 150 days of use. 

Solution:

Let us consider X be the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here, p = 0.05 and  q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability (none will fuse after 150 days of use) is 

P(X = 0) = ^{5}{}{C}_0 \left( \frac{1}{20} \right)^0 \left( \frac{19}{20} \right)^{5 - 0}

= (19/20)5

(ii) Find the probability that out of 5 such bulbs, not more than one will fuse after 150 days of use. 

Solution:



Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95.

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability (not more than 1 will fuse after 150 days of use) is

P(X < 1) = P(X = 0) + P(X = 1)

\left( \frac{19}{20} \right)^5 + 5 C_1 \left( \frac{1}{20} \right)^1 \left( \frac{19}{20} \right)^{5 - 1}

\left( \frac{19}{20} \right)^4 \left\{ \frac{19}{20} + \frac{5}{20} \right\}

= (6/5) (19/20)4

(iii) Find the probability that out of 5 such bulbs more than one will fuse after 150 days of use 

Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability(more than one will fuse after 150 days of use) is

P(X > 1) = 1 – P(X < 1)

= 1 – (6/5) (19/20)4

(iv) Find the probability that out of 5 such bulbs at least one will fuse after 150 days of use. 

Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability(at least one will fuse after 150 days of use) is

P(X > 1) = 1 – P(X = 0)

= 1 – (19/20)5

Question 18. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution:

Let us consider X denotes the number of people that are right-handed in the sample of 10 people.

Now, a binomial distribution X follows with n = 10.

Here p = 90 % = 90/100 = 0.9

And q = 1 – p = 0.1

P(X = r) = 10Cr (0.9)r (0.1)10-r

So, the probability that at most 6 are right – handed is

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

= 1 – {P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}

1 - \sum^{10}_{r = 7}{^{10}{}{C}_r} (0 . 9 )^r (0 . 1 )^{10 - r}

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