WORD PROBLEMS ON BINOMIAL DISTRIBUTION

Problem 1 :

X is a binomial variable such that

2 P(X = 2) = P(X = 3)

the mean of X is known to be 10/3. What would be the probability that X assumes at most the value of 2 ?

a)  16/81    b)  17/81   c)  47/243     d)  46/243

Solution :

2 P(X = 2) = P(X = 3)

The required binomial distribution is,

Problem 2 :

Assuming that one third of the population is tea drinkers and each 1000 enumerators takes a sample of 8 individuals to find out whether they are tea drinkers or not, how many enumerators are expected to report that five or more people are tea drinkers ?

a)  100   b)  95    c)   88    d)  90

Solution :

Number of individuals to be taken = number of trials = 8

Probability of getting tea drinkers p = 1/3

q = 2/3

Problem 3 :

If X ~ B(n, p) what would be the greatest value of the variance of x when n = 16 ?

a)  2      b)  4     c)  8     d)  √5

Solution :

Variance = npq

Variance of X attains its maximum value at p = q = 0.5 and its maximum value is n/4

Here n = 16, then 16/4 = 4.

Problem 4 :

If a random variable X follows binomial distribution with mean as 5 and satisfying the condition 10 P(x = 0) = P(x = 1), what is the value of P(x ≥ 1 / x > 0)?

A)  0.67     b)  0.56     c)  0.99    d) 0.82

Solution :

np = 5

10 P(x = 0) = P(x = 1)

Problem 5 :

It is known that the population of a missile hitting a target is 1/8, what is the probability that out of 10 missiles fired, at least 2 will hit the target ?

a)  0.4258     b)  0.3968     c)  0.5238       d)  0.3611

Solution :

Probability for success p = 1/8, then q = 7/8

n = 10

p(x  ≥ 2) = 1 - p(x < 2)

Problem 6 :

Out of 128 families with 4 children each, how many are expected to have at least one boy and one girl ?

a)  100     b)  105     c)  108     d)  112

Solution :

Sample space = 2⁴

n(S) = 16

Probability of choosing at least one boy and one girl = 1- Probability of choosing all boys and girls

= 1 - (2/16)

= 14/16

= 7/8

In 128 families = 128 (7/8)

= 112

Problem 7 :

In 10 independent rollings of a biased die, the probability that an even number will appear 5 times is twice the probability that an even number will appear 4 times. What is the probability that an even number will appear twice when a dies is rolled 8 times ?

a)  0.0304     b)  0.1243      c)  0.2315      d)  0.1926

Solution :

5P(x = 5) = 2P(x = 4)

Here number of trials with one die = 8