Suppose that 80% of all families own a television set. If 5 families are inervised at random, find the probability that :
(a) three families own a television set.
(b) at least two families own a television set.

Explanation

Since there are 5 families under an interview, n = 5 if they own a television set, we call the outcome a success.
Thus , p = P (success) = 0.8 and q = 1 - p =0.2
Let X : Number of families who owns a televison set out of 5.
Then , `X ~ B(n = 5, p = 0.8)`
The p.m.f of X is given as
`P(X = x) = P(x)`
`""^(5)C_(x)(0.8)^(x) (0.2)^(5-x), X = 0.1,2,3,4,5.`
(a) Probability (three families own a television set)
= `P(X = 3)`
` = ""^(5)C_(3) (0.8)^(3) (0.2)^(2)`
` = (10)(0.512)(0.04)`
` = 0.2048`
(b) Probability (atleast two families own a television set)
`P(X ge 2) = 1 - P(X lt 2)`
` = 1- [P(X = 0) +P(X = 1)]`
`=1[(""^(50)C_(0)(0.8)^(3)(0.2)^(2)+ ""^(5)C_(1) (0.8)^(1)(0.2)^(4)]`
`= 1 - [(1)(1) (0.00032 + 0.0064]`
` = 1- 0.00672`
`=0.99328`