Students can cross-reference their work with OP Malhotra ISC Class 12 Solutions Chapter 20 Theoretical Probability Distribution Ex 20(d) to ensure accuracy.
S Chand Class 12 ICSE Maths Solutions Chapter 20 Theoretical Probability Distribution Ex 20(d)
Question 1.
The mean of a binomial distribution is 20 and standard deviation is 4 . Find out n, p and q.
Answer:
Given mean of binomial distribution =20
⇒ n p = 20 …….(1)
standard deviation of binomial distribution =4
⇒ \(\sqrt{n p q}\) = 4
⇒ n p q = 16 ………(2)
From eqn. (1) and eqn. (2); we have
20 × q = 16
⇒ q = \(\frac{16}{20}\) = \(\frac{4}{5}\)
∴ p = 1 – q = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ from (1); n × \(\frac{1}{5}\) = 20
⇒ n = 100
Question 2.
Determine the binomial distribution whose
(i) mean is 9 and whose Standard deviation is \(\frac{3}{2}\)
(ii) ‘ mean is 10 and starraard deviation is 2 \(\sqrt{2}\).
Answer:
(i) Let the required binomial distribution be (q + p)n
given Mean of binomial distribution =9 ⇒ n p = 9 ……………………(1)
and S.D of binomial distribution = \(\frac{3}{2}\)
⇒ \(\sqrt{n p q}\) = \(\frac{3}{2}\)
⇒ n p q = \(\frac{9}{4}\) …………………(2)
From (1) and (2); we have 9 × q = \(\frac{9}{4}\)
⇒ q = \(\frac{1}{4}\)
∴ p = 1 – q = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∴ from (1); n × \(\frac{3}{4}\) = 9
⇒ n = 12
Thus required binomial distribution is (\(\frac{1}{4}\) + \(\frac{3}{4}\))12
(ii) Let the required binomial distribution is (q + p)n
Given mean of binomial distribution =10 ⇒ n p = 10 ………………..(1)
and given standard deviation of B.D = 2 \(\sqrt{2}\)
⇒ \(\sqrt{n p q}\) = 2 \(\sqrt{2}\)
⇒ n p q = 8 ………………….(2)
From (1) and (2); we have
q = \(\frac{8}{10}\) = \(\frac{4}{5}\)
∴ p = 1 – q = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ from (1); n × \(\frac{1}{5}\) = 10 ⇒ n = 50
Hence the required B.D be (\(\frac{4}{5}\) + \(\frac{1}{5}\))50
Question 3.
If two dice are rolled 12 times, obtain the mean and the variance of the distribution of successes, if getting a total greater than 4 is considered a success.
Answer:
When two dice are rolled
Then S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1), (3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3), (5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
Total no. of outcomes = 62 = 36
Given getting a total greater than 4 is considered as success.
∴ favourable outcomes are
{(1,4),(2,3),(4,1),(3,2),(2,4),(1,5),(3,3),(4,2),(5,1),(1,6),(2,5),(3,4),(4,3),(5,2) (6,1),(2,6),(3,5),(4,4),(5,3),(6,2),(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6)
(6,5),(6,6)}
∴ Total no. of favourable outcomes = 30
Thus p = \(\frac{30}{36}\) = \(\frac{5}{6}\) and
q = 1 – p = 1 – \(\frac{5}{6}\)
= \(\frac{1}{6}\) and n = 12
∴ Mean of binomial distribution = n p = 12 × \(\frac{5}{6}\) = 10 and variance of B.D. = n p q = 12 × \(\frac{5}{6}\) × \(\frac{1}{6}\) = \(\frac{5}{3}\)
Question 4.
The sum of mean and variance of a binomial distribution is \(\frac{35}{16}\) for 5 trials. Find the distribution.
Answer:
Let the required binomial distribution be (q + p)n . given n = 5, sum of mean and variance of B.D. = \(\frac{35}{16}\)
⇒ n p + n p q = \(\frac{35}{16}\)
⇒ 5(p + p q) = \(\frac{35}{16}\)
⇒ p(1 + q) = \(\frac{7}{16}\)
⇒ (1 – q)(1 + q) = \(\frac{7}{16}\)
⇒ 1 – q2 = \(\frac{7}{16}\)
∵ p + q = 1]
∴ = 1 – q = 1 – \(\frac{3}{4}\)
= \(\frac{1}{4}\)
Thus required binomial distribution be (\(\frac{3}{4}\) + \(\frac{1}{4}\))5
Question 5.
The sum of mean and variance of a binomial distribution of 18 trials, is 10 , find the distribution.
Answer:
Let the required binomial distribution be (q + p)n where n = no. of trials = 18
Mean of binomial distribution = n p and variance of B.D. = n p q
Since, it is given that sum of mean and variance of B.D be 10 .
∴ n p + n p q = 10
⇒ 18(p + p q) = 10
⇒(1 – q)(1 + q) = \(\frac{5}{9}\)
[∵ p + q = 1]
⇒ 1 – q2 = \(\frac{5}{9}\)
⇒ q^2 = 1 – \(\frac{5}{9}\) = \(\frac{4}{9}\)
⇒ q = \(\frac{2}{3}\) [∵ q > 0]
∴ p = 1 – q = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Hence the required binomial distribution be (\(\frac{2}{3}\) + \(\frac{1}{3}\))18
Question 6.
(i) The sum of mean and variance of a binomial distribution is 1 5 and the sum of their squares is 117. Find the distribution.
(ii) The difference between the mean and variance of a binomial distribution is 1 and the difference of their squares is 11 . Find the distribution.
Answer:
(i) Let the required binomial distribution be (q + p)n Since mean of B.D = n p and variance of B.D. = n p q
Since the sum of mean and variance of B.D is 15
⇒ n p + n p q = 15
⇒ n p(1 + q) = 15 …………………….(1)
also sum of squares of mean and variance be 117
∴ n2 p2 + n2 p2 q2 = 117
⇒ n2 p2 (1 + q2 ) = 117 ………………..(2)
On squaring eqn. (1); we have
n2 p2 (1 + q)2 = 225 …………………….(3)
On dividing (3) by (2); we have
\(\frac{(1+q)^2}{1+q^2}\) = \(\frac{225}{117}\)
⇒ \(\frac{1+q^2+2 q}{1+q^2}\) = \(\frac{225}{117}\)
⇒ \(\frac{2 q}{1+q^2}\) = \(\frac{225}{117}\) – 1
⇒ \(\frac{2 q}{1+q^2}\) = \(\frac{108}{117}\)
⇒ 108 q2 – 234 q + 108 = 0
⇒ 12 q2 – 26 q + 12 = 0
⇒ 6 q2 – 13 q + 6 = 0
⇒ q = \(\frac{13 \pm \sqrt{169-144}}{12}\)
= \(\frac{13 \pm 5}{12}\)
⇒ q = \(\frac{3}{2}\), \(\frac{2}{3}\)
but 0 < p, q < 1
∴ q = \(\frac{2}{3}\)
Thus p = 1 – q = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
∴ from (1); n × \(\frac{1}{3}\) × (1 + \(\frac{2}{3}\)) = 15
⇒ n × \(\frac{1}{3}\) × \(\frac{5}{3}\) = 15
⇒ n = \(\frac{15 \times 9}{5}\) = 27
Thus the required distribution be (q + p)n
i.e. (\(\frac{2}{3}\) + \(\frac{1}{3}\))27 .
(ii) We know that, mean of binomial distribution = n p and variance of B.D. = n p q according to given condition, we have
andn p – n p q = 1
⇒ n p(1 – q) = 1 ……………………(1)
n2 p2 – (n p q)2 = 11
⇒ n2 p2 (1 – q2 ) = 11 …………………….(2)
squaring eqn. (1); we have
n2 p2 (1 – q)2 = 1 ……………………….(3)
On dividing eqn. (3) by eqn. (2); we have
\(\frac{(1-q)^2}{1-q^2}\) = \(\frac{1}{11}\)
⇒ 11(1 + q2 – 2 q) = 1 – q2
⇒ 12 q2 – 22 q + 10 = 0
⇒ 6 q2 – 11 q + 5 = 0
⇒(q – 1)(6 q – 5) = 0
⇒ q = 1, \(\frac{5}{6}\)
When q = 1
∴ p = 1 – q = 1 – 1 = 0, which is not possible.
Thus q = \(\frac{5}{6}\)
p = 1 – q = 1 – \(\frac{5}{6}\) = \(\frac{1}{6}\)
∴ from (1); n × \(\frac{1}{6}\)(1 – \(\frac{5}{6}\)) = 1
⇒ \(\frac{n}{36}\) = 1
⇒ n = 36
Thus the required binomial distribution be (q + p)n
i.e. (\(\frac{5}{6}\) + \(\frac{1}{6}\))36
Question 7.
(i) Comment on the following : For a binomial distribution, mean = 7 and variance = 11.
(ii) Can the mean of a binomial distribution be less than variance?
Bring out the fallacy, if any, in the statement. “The mean of a binomial distribution is 15 and its standard deviation is 5 .”
Answer:
(i) Given mean of binomial distribution = 7
⇒ n p = 7 …………..(1)
and variance of binomial distribution = 11
⇒ n p q = 11 …………….(2)
From eqn. (1) and eqn. (2); we have
7 q = 11
⇒ q = \(\frac{11}{7}\) > 1 which is impossible [∵ 0 < q < 1]
Hence such a binomial distribution is not possible.
(ii) Since mean of binomial distribution = n p and variance of B.D = n p q
Here, Mean – variance = n p – n p q = n p(1 – q) = n p2 > 0
[∵ p + q = 1]
[∵ n N, p > 0 ⇒ n p2 > 0]
⇒ mean > variance
Thus mean of B.D can’t be less than variance.
Given mean of B.D =15 ⇒ n p = 15 …………..(1)
and S.D of B.D. = 5
⇒ \(\sqrt{n p q}\) = 5 ⇒ n p q = 25 …………….(2)
From eqn. (1) and eqn. (2); we have
15 q = 25 ⇒ q = \(\frac{25}{15}\) = \(\frac{5}{3}\) > 1 [∵ 0 < q < 1]
which is impossible.
So such a binomial distribution is not possible.
Here Mean = 15; Variance = (S.D)2 = 52 = 25
i.e. Mean < Variance.
Question 8.
If the random variable X follows binomial distribution with mean 3 and variance \(\frac{3}{2}\), find P(X < 5).
Answer:
Given mean of binomial distribution = 3 ⇒ n p = 3 ………….(1)
and Variance of B.D. = \(\frac{3}{2}\)
⇒ n p q = \(\frac{3}{2}\) …………(2)
From (1) and (2); we have
3 q = \(\frac{3}{2}\)
⇒ q = \(\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ from (1); n × \(\frac{1}{2}\) = 3
⇒ n = 6
Thus by binomial distribution, we have
P(X = r)
= n Cr pr qn-r
= 6 C2 (\(\frac{1}{2}\))r (\(\frac{1}{2}\))6-r
= 6 Cr (\(\frac{1}{2}\))6
∴ P(X < 5)
= 1 – P(X = 6)
= 1 – 6 C6 (\(\frac{1}{2}\))6
= 1 – \(\frac{1}{64}\) = \(\frac{63}{64}\)
Question 9.
The sum and product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution. (NMOC)
Answer:
We know that, mean of B.D = n p and variance of B.D. = n p q according to given condition, we have
and
n p + n p q = 24
⇒ n p(1 + q) = 24 ………..(1)
(n p)(n p q) = 128
⇒ n2 p2 q = 128 …………….(2)
On squaring eqn. (1); we have
n2 p2 (1 + q)2 = 576 …………..(3)
On dividing eqn. (3) by eqn. (2); we get
\(\frac{n^2 p^2(1+q)^2}{n^2 p^2 q}\) = \(\frac{576}{128}\)
= \(\frac{32 \times 18}{32 \times 4}\) = \(\frac{9}{2}\)
⇒ \(\frac{(1+q)^2}{q}\) = \(\frac{9}{2}\)
⇒ 2(1+q)2 = 9 q
⇒ 2 q2 + 4 q + 2 = 9 q
⇒ 2 q2 – 5 q + 2 = 0
⇒ (q – 2)(2 q – 1) = 0
⇒ q = 2, \(\frac{1}{2}\)
since 0 < q < 1
∴ q = 2 is impossible
Thus, q = \(\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ from (1); n × \(\frac{1}{2}\)(1 + \(\frac{1}{2}\)) = 24
⇒ n × \(\frac{1}{2}\) × \(\frac{3}{2}\) = 24
⇒ n = \(\frac{24 \times 4}{3}\) = 32
Hence, the required binomial distribution be (q + p)n
i.e. (\(\frac{1}{2}\) + \(\frac{1}{2}\))32
Question 10.
If the mean of a binomial distribution is 24 and its standard deviation is 4 , calculate the number of observations and the relative frequency of occurrence of the event.
Answer:
Let n be the number of observations
Since mean of B.D = n p
∴ n p = 24 …………..(1)
and S.D of B.D. = \(\sqrt{n p q}\)
∴ \(\sqrt{n p q}\) = 4
⇒ n p q = 16 …………..(2)
From (1) and (2); we have
24 q = 16
⇒ q = \(\frac{16}{24}\) = \(\frac{2}{3}\)
∴ p = 1 – q = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
∴ from (1); n × \(\frac{1}{3}\) = 24
⇒ n = 72
Thus, the required no. of observations = 72
Question 11.
The mean number of success of a binomial distribution (p + q)n is 240 , where p is the probability of success. The standard deviation is 12 . Calculate the values of n, p and q.
Answer:
Given mean of binomial distribution = 240
⇒ n p = 240 ………….(1)
and standard deviation of B.D. = 12
⇒ \(\sqrt{n p q}\) = 12
⇒ n p q = 144 ……………………(2)
On dividing eqn. (2) by eqn. (1); we have
\(\frac{n p q}{n p}\)
= \(\frac{144}{240}\)
⇒ q = \(\frac{6}{10}\) = \(\frac{3}{5}\)
∴ p = 1 – q = 1 – \(\frac{3}{5}\)
= \(\frac{2}{5}\)
∴ from (1); we have
n × \(\frac{2}{5}\) = 240
⇒ n = \(\frac{240 \times 5}{2}\) = 600
Question 12.
If the mean of a binomial distribution is 960 and its standard deviation is 24 , calculate the number of observations; and the probability of occurrence of the event.
Answer:
Let the required no. of observation of binomial distribution be n.
Then n p = 960 …………(1)
and \(\sqrt{n p q}\) = 24
⇒ n p q = 576 ………………..(2)
On dividing eqn. (2) by eqn. (1); we have
\(\frac{n p q}{n p}\) = \(\frac{576}{960}\)
⇒ q = \(\frac{6}{10}\) = \(\frac{3}{5}\)
∴ p = 1 – q = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
∴ from (1); n × \(\frac{2}{5}\) = 960
⇒ n = \(\frac{960 \times 5}{2}\) = 2400
Thus, the probability of an occurrence of an event = p = \(\frac{2}{5}\)
Question 13.
Acertain brand of razor blades is sold in packets of 5 . The following is the frequency distribution of 100 packets according to the number of faulty blades in them.
| No.of faulty blades | No.of packets |
| 0
1 2 3 |
80
17 2 1 |
| 4 or more | 0 |
Find the number of faulty blades per packet. Assuming that the distribution is binomial, estimate the probability that a blade taken at random from any packet will be faulty.
Answer:
∴ Mean number of faulty blades per packet = \(\frac{0 \times 80+1 \times 17+2 \times 2+3 \times 1}{100}\)
= \(\frac{24}{100}\) = 0.24
Since the distribution is binomial distribution
∴ n p = 0.24
Also n = 5
∴ p = \(\frac{0.24}{5}\)
= \(\frac{24}{500}\) = 0.048
Thus, the required no. of faulty blades per packet = 0.048
Question 14.
A coin is tossed (i) 10 times, (ii) 100 times, (iii) 1000 times. Calculate in each case the expected number of heads and the standard deviation. Which would surprise you more 3 heads and 7 tails in 10 tosses or 300 heads and 700 tails in 1000 tosses? Justify your answer mathematically.
(ISC)
Answer:
(a) (i) Here n = 10, p = prob. of getting head in single toss of coin
⇒ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ Mean = n p = 10 × \(\frac{1}{2}\) = 5;
and
S.D = \(\sqrt{n p q}\) = \(\sqrt{10 \times \frac{1}{2} \times \frac{1}{2}}\)
= \(\sqrt{\frac{5}{2}}\)
= 1.58
(ii) Here n = 100, p = q = \(\frac{1}{2}\)
∴ Mean = n p = 100 × \(\frac{1}{2}\) = 50 ；
and
S.D = \(\sqrt{n p q}\)
= \(\sqrt{100 \times \frac{1}{4}}\)
= \(\sqrt{25}\) = 5
(iii) Here n = 1000; p = q = \frac{1}{2} ;
Mean = n p = 1000 × \(\frac{1}{2}\) = 500
and standard deviation = \(\sqrt{n p q}\) = \(\sqrt{1000 \times \frac{1}{2} \times \frac{1}{2}}\)
= \(\sqrt{250}\)
= 5 \(\sqrt{10}\) = 15.8
(b) (i) Here actual no. of heads in 10 tosses = 3
expected no. of heads = 5
∴ difference = 5 – 3 = 2
Here
σ = \(\sqrt{n p q}\)
= \(\sqrt{10 \times \frac{1}{2} \times \frac{1}{2}}\)
= \(\sqrt{\frac{5}{2}}\) = 1.58
and
\(\frac{\text { difference }}{sigma}\) = \(\frac{2}{1.58}\) = 1.3
⇒ diff. = 1.3 σ < 2 σ Here coin is not biased.
(ii) Here, actual no. of heads in 1000 tosses = 300 Expected no. of heads = 500
∴ difference = 500 – 300 = 200
σ = \(\sqrt{n p q}\)
= \(\sqrt{1000 \times \frac{1}{2} \times \frac{1}{2}}\) = \(\sqrt{250}\) = 15.8
Here \(\frac{\text { difference }}{sigma}\) = \(\frac{200}{15.8}\) = 12.7 and difference =12.7 σ > 3 σ
Have 300 heads and 700 tails in 1000 tosses would surprise more. Here coin is biased.
Examples:
Question 1.
In a large collection of bulbs, 3 out of 5 are daffodils and the rest are tulips. If they are planted at random, calculate the probability that in a row of five plants (i) all are daffodils, (ii) at least four are daffodils.
Answer:
Let p= probability that daffodils are planted = \(\frac{3}{5}\)
∴ q = 1 – p = 1 – \(\frac{3}{5}\)
= \(\frac{2}{5}\)
Let X be the random variable devotes the no. of daffodils planted out of 5 plants and all the trials are independent of each other
∴ X is a binomial variate with two parameters n = 5 and p = \(\frac{3}{5}\) Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 5 Cr (\(\frac{3}{5}\))r (\(\frac{2}{5}\))5-r ;
r = 0,1,2,3,4,5
(i) Required probability = P(X = 5) = 5 C5 (\(\frac{3}{5}\))5 (\(\frac{2}{5}\))0
= \(\frac{3^5}{5^5}\)
= \(\frac{243}{3125}\)
(ii) Required probability = P(X > 4) = P(X = 4) + P(X = 5)
= 5 C4 (\(\frac{3}{5}\))4 (\(\frac{2}{5}\)) + 5 C5 (\(\frac{3}{5}\))5 (\(\frac{2}{5}\))0
= \(\frac{1}{3125}\)[5 × 81 × 2 + 243]
= \(\frac{1053}{3125}\)
Question 2.
Mohan and Sohan play 12 games of chess. Mohan wins 6 games and Sohan 4 games. 2 games end in a draw. They agree to play 3 more games. Calculate the probability that out of these 3 games, 2 games end in, a draw.
Answer:
Let p = probability that game end in draw = \(\frac{2}{12}\) = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Let X be the random variable denotes the no. of draw games out of 3 .
Since all the three trials are independent of each other.
∴ X is a binomial variate with parameters n = 3 and p = \(\frac{1}{6}\)
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 3 Cr (\(\frac{1}{6}\))r (\(\frac{5}{6}\))3-r ;
r = 0, 1, 2, 3
∴ Required probability = P(X = 2)
= 3 C2 (\(\frac{1}{6}\))2 (\(\frac{5}{6}\))
= \(\frac{15}{216}\)
= \(\frac{5}{72}\)
Question 3.
On an average, out of 12 games of chess played by A and B, A wins 6, B wins 4 and 2 games end in a tie. A and B play a tournament of 3 games. Calculate the probability that
(i) at least 2 games end in a tie.
(ii) A and B win alternate games, no games end in a tie.
Answer:
Let p= probability that game end in tie = \(\frac{2}{12}\) = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Let X be the random variable denote the no. of tie games out of 3 .
Since all the three trials are independent of each other.
∴ X is a binomial variate with two parameters n = 3 and p = \(\frac{1}{6}\) Thus by binomial distribution, we have
P(X = r) = nCr pr qn-r
= 3Cr (\(\frac{1}{6}\))r (\(\frac{5}{6}\))3-r
(i) Required probability = P(X > 2) = P(X = 2) + P(X = 3)
= 3C2 (\(\frac{1}{6}\))2 (\(\frac{5}{6}\)) + 3C3(\(\frac{1}{6}\))3
= \(\frac{3 \times 5}{216}\) + \(\frac{1}{216}\)
= \(\frac{16}{216}\)
= \(\frac{2}{27}\)
(ii) Required probability = P(A wins ) P(B wins ) P(A wins ) + P(B wins ) P(A wins ) P(B wins )
Since,
P(A wins ) = \(\frac{6}{12}\)
= \(\frac{1}{2}\) and P(B wins)
= \(\frac{4}{12}\)
= \(\frac{1}{3}\)
∴ from eqn. (1); we have
∴ Required probability = \(\frac{1}{2}\) × \(\frac{1}{3}\) × \(\frac{1}{2}\) + \(\frac{1}{3}\) × \(\frac{1}{2}\) × \(\frac{1}{3}\)
= \(\frac{1}{12}\) + \(\frac{1}{18}\)
= \(\frac{3+2}{36}\)
= \(\frac{5}{36}\)
Question 4.
The probability density function y of a continuous variable x is given by
y = \(\frac{x}{k}\), 0 < x < 2 and y = 0
for all other values of x. Calculate the value of k and the probability that x<1.
Answer:
Given y = { \(\frac{x}{k}\);
0 < x < 2 0 ;
other values of x.
Since x be continuous random variable and y be the probability density function
Thus 0∫2 y d x = 1
⇒ 0∫2 \(\frac{x}{k}\) d x = 1
⇒ \(\frac{1}{k}\) \(\frac{x^2}{2}\)20 = 1
⇒ \(\frac{2}{k}\) = 1
⇒ k = 2
Thus, P(x < 1) = 0∫1 y d x
= 0∫1 \(\frac{x}{2}\) d x
= \(\frac{x^2}{4}\)10
= \(\frac{1}{4}\)
Question 5.
A pair of dice is thrown 5 times. If getting a doublet is considered a success, find the probability of 2 successes.
Answer:
When a pair of dice is thrown
Then total no. of outcomes = 62 = 36
For getting doublets, favourable outcomes are {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
∴ Total no. of favourable outcomes = 6
Let p = prob. of getting doublet = \(\frac{6}{36}\)
= \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Let X be the random variable denote the no. of doublets obtained with pair of dice throwing 5 times.
Since all the five trials are independent of each other.
∴ X is a binomial variate with parameters n = 5 and p
= \(\frac{1}{6}\)
Thus by binomial distribution; we have
P(X = r) = n Cr pr qn-r
= 5 Cr (\(\frac{1}{6}\))r (\(\frac{5}{6}\))5-r
∴ Required probability = P(X = 2)
= 5 C2 (\(\frac{1}{6}\))2 (\(\frac{5}{6}\))3
= \(\frac{5 \times 4}{2}\) × \(\frac{125}{6^5}\)
= \(\frac{625}{3888}\)
Question 6.
Assume, that on an average, 1 telephone out of 10 is busy, six telephone numbers are randomly selected and called. Find the probability that four of them will be busy.
Answer:
Let p= prob. that telephone number is busy = \(\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\)
Let X be the random variable denotes the no. of telephone numbers that are busy out of 6 randomly selected telephone numbers and all trials are independent of each other.
∴ X is a binomial variate with two parameters p = \(\frac{1}{10}\) and n = 6
Thus by binomial distribution, we have
P(X = r) = nCr pr qn-r
= 6Cr (\(\frac{1}{10}\))r (\(\frac{9}{10}\))6-r
∴ Required probability = P(X = 4)
= 6C4 ( \(\frac{1}{10}\))4 ( \(\frac{9}{10}\))2
= \(\frac{6 \times 5}{2}\) × \(\frac{81}{10^6}\)
= \(\frac{1215}{1000000}\)
= 0.001215
Question 7.
Assume that half the population are consumers of chocolate, so that the chance of an individual being a consumer is \(\frac{1}{2}\). If 100 investigators interviewed 10 individuals each to see whether they are consumers, how many investigators would you expect to report that 3 individuals or less are consumers.
Answer:
Let p = prob. that individual is a chocolate consumer = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)
Let X be the random variable denotes the number of individuals whose are chocolate consumer out of 10 .
Thus X is binomial variate with two parameters n = 10 and p = \(\frac{1}{2}\).
Since all trials are independent of each other.
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 10 Cr(\(\frac{1}{2}\))r (\(\frac{1}{2}\))10-r
= 10 Cr(\(\frac{1}{2}\))10
∴ Required probability = P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 10 C0(\(\frac{1}{2}\))10 + 10 C1(\(\frac{1}{2}\))10 + 10 C2(\(\frac{1}{2}\))10
+ 10 C3(\(\frac{1}{2}\))10
= [1 + 10 + \(\frac{10 \times 9}{2}\) + \(\frac{10 \times 9 \times 8}{6}\)] \(\frac{1}{2^{10}}\)
= \(\frac{(1+10+45+120)}{1024}\)
= \(\frac{176}{1024}\)
= 0.172
Thus the required number of investigators who will report that there are three or less than three are individuals w ho are consum ers = N × P(X > 3) = 100 × 0.172 = 17.2 i.e. 17 persons only.
Question 8.
A and B play a game in which A ‘s chance of winning the game is \(\frac{3}{5}\). In a series of 6 games, find the probability that A will win at least 4 games.
Answer:
Let p= probability of A’s winning = \(\frac{3}{5}\)
∴ q = 1 – p = 1 – \(\frac{3}{5}\)
= \(\frac{2}{5}\); n = 6
Let X be the random variable denote the number of games win by A out of 6 games played.
∴ X is a binomial variate with two parameters n = 6 and p = \(\frac{3}{5}\)
Since all trials are independent of each other.
Thus by binomial distribution, we have
P(X = r) = n CR pr qn-r
= 6 CR (\(\frac{3}{5}\))r (\(\frac{2}{5}\))6-r
∴ Required probability = P(X > 4)
= P(X = 4) + P(X = 5) + P(X = 6)
= 6 C4 (\(\frac{3}{5}\))4 (\(\frac{2}{5}\))2 + 6 C5(\(\frac{3}{5}\))5 (\(\frac{2}{5}\)) + 6 C6(\(\frac{3}{5}\))6
= \(\frac{3^4}{5^6}\)[\(\frac{6 \times 5}{2}\) × 4 + 6 × 3 × 2 + 9]
= \(\frac{3^4}{5^6}\)[60 + 36 + 9]
= \(\frac{81 \times 105}{5^6}\)
= \(\frac{81 \times 21}{5^5}\)
= \(\frac{1701}{3125}\)
Question 9.
A die is tossed three times. Getting a ‘ 3 ‘ or a ‘ 5 ‘ is considered a success. Find the probability of at least two successes.
Answer:
Let p= probability of success = prob. of getting 3 or 5 in single toss of die = \(\frac{2}{6}\)
= \(\frac{1}{3}\)
∴ q = 1 – p = 1 – \(\frac{1}{3}\)
= \(\frac{1}{3}\)
Let X denotes the number of successes in 3 throws and all trials are independent of each other.
∴ X is a binomial variate with two parameters n = 3 and p = \(\frac{1}{3}\)
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 3 Cr (\(\frac{1}{3}\))r (\(\frac{2}{3}\))3-r
∴ required probability = P(X > 2)
= P(X = 2) + P(X = 3)
= 3 C2 (\(\frac{1}{3}\))2 (\(\frac{2}{3}\)) +
3 C3 (\(\frac{1}{3}\))3 (\(\frac{2}{3}\))2
= \(\frac{3 \times 2}{27}\) + \(\frac{1}{27}\)
= \(\frac{7}{27}\)
Question 10.
A coin is tossed 5 times. What is the probability of getting at least three heads?
Answer:
Let p= probability of getting head in single toss of coin = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); n = 5
Let X be the random variable denotes the no. of heads in 5 tosses.
Also all trials are independent of each other.
∴ X is a binomial variate and with two parameters n = 5 and p = \(\frac{1}{2}\)
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 5 Cr (\(\frac{1}{2}\))r (\(\frac{1}{2}\))5-r
= 5 Cr (\(\frac{1}{2}\))5
∴ Required probability
= P(X > 3)
= P(X = 3) + P(X = 4) + P(X = 5)
= 5 C3 (\(\frac{1}{2}\))5 + 5 C4(\(\frac{1}{2}\))5
+ 5 C5 (\(\frac{1}{2}\))5
= (\(\frac{1}{2}\))5 [\(\frac{5 \times 4}{2}\) + 5 + 1]
= \(\frac{16}{32}\)
= \(\frac{1}{2}\)
Question 11.
Four dice are thrown simultaneously. If the occurrence of an odd number in a single die is considered a success, find the probability of at most 2 successes.
Answer:
Let p = probability of success i.e. prob. of getting an odd number in single throw of dice = \(\frac{3}{6}\)
= \(\frac{1}{2}\)
∴ q = 1 – p
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\); n = 4
Let X be the random variable denotes the number of successes in four throws. All trials are independent of each other.
Thus X is binomial variate with two parameters n = 4, p = \(\frac{1}{2}\)
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 4 Cr (\(\frac{1}{2}\))r (\(\frac{1}{2}\))4-r
= 4 Cr (\(\frac{1}{2}\))4
∴ Required probability = P(X > 2)
= P(X = 0) + P(X = 1) + P(X = 3)
= 4 C0 (\(\frac{1}{2}\))4 + 4 C1 (\(\frac{1}{2}\))4 + 4 C2 (\(\frac{1}{2}\))4
= (\(\frac{1}{2}\))4 [1 + 4 + \(\frac{4 \times 3}{2}\)]
= \(\frac{11}{16}\)
Question 12.
Assuming that on an average one tele- phone out of ten is busy, seven telephone numbers are randomly selected as called. Find the probability that three of them will be busy.
Answer:
Let p = prob. of telephone number be busy = \(\frac{1}{10}\)
∴ q = 1 – p
= 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\); n = 7
Let X be the random variable denote the no. of telephone numbers which are busy out of 7 randomly selected telephone numbers and all trials are independent of each other.
∴ X is a binomial variate with parameters n = 7 and p =\( \frac{1}{10}\)
Thus binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 7 Cr (\(\frac{1}{10}\))r (\(\frac{9}{10}\))7-r
∴ Required probability = P(X = 3)
= 7 C3 (\(\frac{1}{10}\))3 (\(\frac{9}{10}\))4
= \(\frac{7 \times 6 \times 5}{6}\) × \(\frac{9^4}{10^7}\)
= \(\frac{229635}{10^7}\)
Question 13.
In a binomial distribution, the sum of its mean and the variance is 1.8 . Find the probability of two successes if the event was conducted 5 times.
Answer:
Let p be the probability of success and n = 5
given sum of mean and variance = 1.8
⇒ n p + n p q = 1.8
⇒ n p(1 + q) = 1.8
⇒ 5(1 – q)(1 + q) = 1.8
[∵ p + q = 1]
⇒ 1 – q2 = \(\frac{1.8}{5}\) = \(frac{18}{50}\) = \(\frac{9}{25}\)
⇒ q2 = 1 – \(\frac{9}{25}\) = \(\frac{16}{25}\)
⇒ q = \(\frac{4}{5}\)
[∵ 0 < q < 1]
∴ p = 1 – q = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Thus by binomial distribution, we have
P(X = r) = 2Cr pr qn-r
= 5Cr (\(\frac{1}{5}\))r (\(\frac{4}{5}\))5-r
∴ Required probability = P(X = 2)
= 5C2 (\(\frac{1}{5}\))2 (\(\frac{4}{5}\))3
= \(\frac{5 \times 4}{2}\) × \(\frac{64}{5^5}\)
= \(\frac{640}{3125}\)
= \(\frac{128}{625}\)
Question 14.
Eight coins are thrown simultaneously.
(i) Show that the probability of getting at least 6 heads is \(\frac{37}{256}\).
(ii) What is the probability of getting at least 3 heads ?
Answer:
Let p = probability of geting a head = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Let X be the random variable denotes the number of heads obtained in 8 tosses and all the 8 trials are independent of each other.
∴ X is binomial variate with parameters n = 8 and p = \(\frac{1}{2}\)
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 8Cr (\(\frac{1}{2}\))r (\(\frac{1}{2}\))8-r
= 8Cr (\(\frac{1}{2}\))8
(i) ∴ Required probability = P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)
= 8 C6(\(\frac{1}{2}\))8 + 8 C7(\(\frac{1}{2}\))8 + 8 C8(\(\frac{1}{2}\))8
= \(\frac{1}{2^8}\)[\(\frac{8 \times 7}{2}\) + 8 + 1]
= \(\frac{37}{256}\)
(ii) Required probability = P(X < 3)
= 1 – P(X = 0) – P(X = 1) – P(X = 2)
= 1 -[8 C0 (\(\frac{1}{2}\))8 + 8 C1(\(\frac{1}{2}\))8 + 8 C2(\(\frac{1}{2}\))8]
= 1 – \(\frac{1}{2^8}\)[1 + 8 + 28]
= 1 – \(\frac{37}{256}\)
= \(\frac{219}{256}\)
Question 15.
The mean and variance of a binomial distribution are 4 and 2 respectively. Find the probability of at least 6 successes.
Answer:
We know that, mean of B.D. = n p and variance of B.D = n p q
Then
n p = 4 …………(1)
and = 2 ……………(2)
From eqn. (1) and cqu. (2); we have
4 × q = 2
⇒ q = \(\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ from (1); n × \(\frac{1}{2}\) = 4
⇒ n = 8
Thus, by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 8 Cr (\(\frac{1}{2}\))r (\(\frac{1}{2}\))8-r
= 8 Cr (\(\frac{1}{2}\))8
∴ Required probability = P(X > 6)
= P(X = 6) + P(X = 7) + P(X = 8)
= 8 C6 (\(\frac{1}{2}\))8 + 8 C7(\(\frac{1}{2}\))8 + 8 C8(\(\frac{1}{2}\))8
= \(\frac{1}{2^8}\)[\(\frac{8 \times 7}{2}\) + 8 + 1]
= \(\frac{37}{256}\)
Question 16.
Five bad eggs aretaixed with 10 good ones. If three eggs are drawn one by one with replacement, find the probability distribution of the number of good eggs drawn.
Answer:
Given number of bad eggs = 5
number of good eggs = 10
∴ Total no. of eggs = 5 + 10 = 15
Let p = probability of getting a good egg = \(\frac{10}{15}\)
= \(\frac{2}{3}\)
∴ q = 1 – p = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Let X be the random variable denote the no. of good eggs in three draws.
∴ X can take values 0,1,2,3.
P(X = 0) = P(no good egg ) = q q
q = \(\frac{1}{3}\) × \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{27}\)
P(X = 1) = P(1 good egg and 2 bad ones )
= p q q + q p q + q q p
= \(\frac{2}{3}\) × \(\frac{1}{3}\) × \(\frac{1}{3}\) + \(\frac{1}{3}\) × \(\frac{2}{3}\) × \(\frac{1}{3}\) + \(\frac{1}{3}\) × \(\frac{1}{3}\) × \(\frac{2}{3}\)
= \(\frac{6}{27}\)
= \(\frac{2}{9}\)
P(X = 2) = P(2 good eggs and one bad egg)
= p p q + p q p + q p p
= \(\frac{2}{3}\) × \(\frac{2}{3}\) × \(\frac{1}{3}\) + \(\frac{2}{3}\) × \(\frac{1}{3}\) × \(\frac{2}{3}\) + \(\frac{1}{3}\) × \(\frac{2}{3}\) × \(\frac{2}{3}\)
= \(\frac{12}{27}\)
= \(\frac{4}{9}\)
P(X = 3) = P(3 good eggs) = p p
p = \(\frac{2}{3}\) × \(\frac{2}{3}\) × \(\frac{2}{3}\)
= \(\frac{8}{27}\)
Then the probability distribution of X is given by
| X | 0 | 1 | 2 | 3 |
| P(X) | \(\frac{1}{27}\) | \(\frac{2}{9}\) | \(\frac{4}{9}\) | \(\frac{8}{27}\) |
Question 17.
A box contains 4 red and 5 black marbles. Find the probability distribution of the red marbles in a random draw of three marbles. Also, find the mean and standard deviation of the distribution.
Answer:
Let p = probability of drawing a red marble in a bag
= \(\frac{4}{9}\)
∴ q = 1 – p = 1 – \(\frac{4}{9}\)
= \(\frac{5}{9}\)
Let X denote the no. of red marbles drawn in a draw of three marbles thus X can take values 0,1 , 2,3 .
P(X = 0) = prob. of drawing no red marble
= \(\frac{\frac{5 \times 4}{2}}{\frac{9 \times 8 \times 7}{6}}\)
= \(\frac{5}{42}\)
P(X = 1) = prob. of drawing one red marble and 2 black marables
= \(\frac{4 \times \frac{5 \times 4}{2}}{\frac{9 \times 8 \times 7}{6}}\)
= \(\frac{40}{84}\)
= \(\frac{10}{21}\)
P(X = 2) = prob. of drawing two red and 1 black marble
= \(\frac{\frac{4 \times 3}{2} \times 5}{\frac{9 \times 8 \times 7}{6}}\)
= \(\frac{30}{84}\)
= \(\frac{15}{42}\)
= \(\frac{5}{14}\)
P(X = 3) = prob. of drawing three red balls
= \(\frac{4}{84}\)
= \(\frac{1}{21}\)
Thus the probability distribution of X is given below :
| X | 0 | 1 | 2 | 3 |
| P(X) | \(\frac{5}{42}\) | \(\frac{10}{21}\) | \(\frac{5}{14}\) | \(\frac{1}{21}\) |
The table of values for computation of mean and variance is given as under
| Xi | Pi | PiXi | Pi2Xi2 |
| 0 | \(\frac{5}{42}\) | 0 | 0 |
| 1 | \(\frac{10}{21}\) | \(\frac{10}{21}\) | \(\frac{10}{21}\) |
| 2 | \(\frac{5}{14}\) | \(\frac{10}{14}\) | \(\frac{20}{14}\) |
| 3 | \(\frac{1}{21}\) | \(\frac{3}{21}\) | \(\frac{9}{21}\) |
| PiXi = \(\frac{56}{42}\) | Pi2Xi2= \(\frac{98}{42}\) |
∴ Mean = µ = ∑pi xi = \(\frac{56}{42}\) = \(\frac{4}{3}\)
∴ Variance = ∑2 = ∑pi xi2 – µ2
= \(\frac{98}{42}\) – (\(\frac{4}{3}\))2
= \(\frac{7}{3}\) – \(\frac{16}{9}\)
= \(\frac{21-16}{9}\)
= \(\frac{5}{9}\)
Thus S.D = ∑
= \(\sqrt{\frac{5}{9}}\)
= \(\frac{1}{3}\) \(\sqrt{5}\)
Thus
S.D
= ∑ = \(\sqrt{\frac{5}{9}}\)
= \(\frac{1}{3}\) \(\sqrt{5}\)
Question 18.
The probability that a bulb produced by a factory will fuse in 100 days of use is 0.05 . Find the probability that out of 5 such bulbs after 100 days of use:
(i) None
(ii) Not more than one
(iii) More than one
(iv) At least one will fuse.
Answer:
Let p = prob. that bulb produced by a factory will fuse after 100 days of use = 0.05 = \(\frac{1}{20}\)
∴ q = 1 – p = 1 – 0.05 = 0.95
= \(\frac{95}{100}\)
= \(\frac{19}{20}\)
Let X be the random variable denotes the no. of bulbs which is fused after 100 days of use out of 5 such bulbs.
Since all trials are independent of each other.
∴ X is binomial variate with parameters n = 5 and
p = \(\frac{1}{20}\) Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 5 Cr (\(\frac{1}{20}\))r (\(\frac{19}{20}\))5-r
(i) Required probability = P(X = 0)
= 5 C0(\(\frac{1}{20}\))0 (\(\frac{19}{20}\))5
= (\(\frac{19}{20}\))5
(ii) Required probability = P(X < 1) = P(X = 0) + P(X = 1)
= 5 C0(\(\frac{1}{20}\))0 (\(\frac{19}{20}\))5 + 5 C1(\(\frac{1}{20}\)) (\(\frac{19}{20}\))4
= (\(\frac{19}{20}\))4
= \(\frac{1\times19+5}{20}\)4
= \(\frac{6}{5}\)(\(\frac{19}{20}\))4
(iii) Required probability = P(X > ) = 1 – P(X = 0) – P(X = 1)
= 1 – \(\frac{6}{5}\)(\(\frac{19}{20}\))5
[using part (ii)]
(iv) Required probability = P(X > 1) = 1 – P(X – 0)
= 1 – 5 C0 (\(\frac{1}{20}\))0 (\(\frac{19}{20}\))5
= 1 – (\(\frac{19}{20}\))5
Question 19.
If the sum and product of the mean and variance of a Binomial distribution are 1.8 and 0.8 respectively, find the probability distribution and the probability of at least one success.
Answer:
In binomial distribution, mean = n p and variance = n p q according to given condition, we have
np + npq = 1.8
⇒ np(1 + q) = 1.8 …………… (1)
and (np)(npq) = 0.8
⇒ n2 p2 q = 0.8 …………..(2)
On squaring eqn. (1); we have
n2 p2 (1+q)2 = (1.8)2 …………..(3)
On dividing eqn. (3) by eqn. (2); we have
\(\frac{n^2 p^2(1+q)^2}{n^2 p^2 q}\) = \(\frac{(1.8)^2}{0.8}\)
⇒ \(\frac{(1+q)^2}{q}\) = \(\frac{324 \times 10}{100 \times 8}\)
⇒ \(\frac{(1+q)^2}{q}\) = \(\frac{324}{80}\)
= \(\frac{81}{20}\)
⇒ 20 q2 + 40 q + 20 = 81 q
⇒ 20 q2 – 41 q + 20 = 0
∴ q = \(\frac{41 \pm \sqrt{(41)^2-1600}}{40}\)
= \(\frac{41 \pm 9}{40}\)
= \(\frac{5}{4}\), \(\frac{32}{40}\)
i.e. \(\frac{4}{5}\)
since 0 < q < 1
∴ q = \(\frac{4}{5}\)
p = 1 – q = 1 – \(\frac{4}{5}\)
= \(\frac{1}{5}\)
∴ from (1); n × \(\frac{1}{5}\) (1 + \(\frac{4}{5}\)) = 1.8
⇒ \(\frac{n}{5}\) × \(\frac{9}{5}\) = 1.8
⇒ n = 0.2 × 25 = 5
Thus by binomial distribution, P(X = r)
= n Cr pr qn-r
⇒ P(X = r)
= Cr(\(\frac{1}{5}\))r (\(\frac{4}{5}\))5-r
∴ Required probability = P(X > 1) = 1 – P(X = 0)
= 1 – 5C0(\(\frac{1}{5}\))0 (\(\frac{4}{5}\))5
= 1 – \(\frac{4^5}{5^5}\)
Question 20.
On dialling certian telephone numbers, assume that on an average one telephone number out of five is busy. Ten telephone numbers are randomly selected and dialled. Find the probability that at least three of them will be busy.
Answer:
Let p = prob. that one telephone number is busy = \(\frac{1}{5}\)
∴ q = 1 – p = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Let X be the random variable denotes the no. of telephone numbers which are busy out of 10 chosen telephone numbers and all trials are independent of each other.
∴ X is binomial variate with parameters p = \(\frac{1}{5}\) and n = 10
Thus by binomial distribution, we have
P(X = r) = n Cr pr qn-r
= 10 Cr (\(\frac{1}{5}\))r (\(\frac{4}{5}\))10-r
∴ Required probability = P(X > 3)
= 1 – P(X = 0) – P(X = 1) – P(X = 2)
= 1 – 10 C0 (\(\frac{1}{5}\))0 (\(\frac{4}{5}\))10 – 10 C1\(\frac{1}{5}\)(\(\frac{4}{5}\))9
– 10 C2(\(\frac{1}{5}\))2 (\(\frac{4}{5}\))8
= 1 – (\(\frac{4}{5}\))8 [\(\frac{16}{25}\) + \(\frac{10 \times 4}{25}\) + \(\frac{10 \times 9}{2}\) × \(\frac{1}{25}\)]
= 1 – \(\frac{101}{25}\)(\(\frac{4}{5}\))8
Question 21.
Five dice are thrown simultaneously. If the occurence of an odd number is a single die is considered a success, find the probability of maximum successes.
Answer:
Let p = probability of success
i . e. prob. of finding odd number with single dice
= \(\frac{3}{6}\)
= \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Let X be the random variable denote the no. of success in 5 and all the trials are independent of each other throws of die.
∴ X is a binomial variate with parameters n = 5 and p = \(\frac{1}{2}\)
Thus by binomial distribution, we have
P(X = r) = nCr pr qn-r
= 5Cr(\(\frac{1}{2}\))r(\(\frac{1}{2}\))5-r
= 5Cr(\(\frac{1}{2}\))5
∴ Required prob. of maximum successes (atmost 3 successes) = P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 5C0(\(\frac{1}{2}\))5 + 5C1(\(\frac{1}{2}\))5 + 5C2(\(\frac{1}{2}\))5 +
5C3(\(\frac{1}{2}\))5
= (\(\frac{1}{2}\))5[1 + 5 + 10 + 10]
= \(\frac{26}{32}\)
= \(\frac{13}{16}\)
Question 22.
The difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution.
Answer:
We know that, mean of binomial distribution = n p and variance of B.D. = n p q according to given condition, we have
and
n p – n p q = 1 ⇒ n p(1 – q) = 1 ……………(1)
n2 p2 – (n p q)2 = 11
⇒ n2 p2 (1 – q2 ) = 11 ……………………..(2)
squaring eqn. (1); we have
n2 p2 (1 – q)2 = 1 …………………….(3)
On dividing eqn. (3) by eqn. (2); we have
\(\frac{(1-q)^2}{1-q^2}\) = \(\frac{1}{11}\)
⇒ 11(1 + q2 – 2 q) = 1 – q2
⇒ 12 q2 – 22 q + 10 = 0
⇒ 6 q2 – 11 q + 5 = 0
⇒ (q – 1)(6 q – 5) = 0
⇒ q = 1, \(\frac{5}{6}\)
⇒When q = 1 ⇒ p = 1 – q = 1 – 1 = 0,
which is not possible.
⇒Thus q = \(\frac{5}{6}\)
∴ p = 1 – q = 1 – \(\frac{5}{6}\) = \(\frac{1}{6}\)
∴ from (1); n × \(\frac{1}{6}\)(1 – \(\frac{5}{6}\)) = 1
⇒ \(\frac{n}{36}\) = 1 ⇒ n = 36
Thus the required binomial distribution be (q + p)n
i.e. (\(\frac{5}{6}\) + \(\frac{1}{6}\))36