Question

Consider that 80 households purchased a television. The customers were surveyed. Results found that 64 households were satisfied with their purchase and 16 households were dissatisfied. Suppose 2 households are randomly selected from the 80 households. Find the probability that both households are dissatisfied with their purchase. Round to four decimal places.

Define A = first household selected is dissatisfied

Define B = second household selected is dissatisfied

Answer 1

Let ,

A = first household selected is dissatisfied.

B = second household selected is dissatisfied.

Total  households = 80 (  64 households were satisfied + 16 households were dissatisfied )

We have to find P( A and B )

P( A ) = 16/80 = 0.2

P( B | A ) -----> Probability of event B given that event A already occurred.

P( B | A ) = 15/79

So, P( A and B ) = P( A ) x P( B |A )

P( A and B ) = ( 16/80) x ( 15/79 )

P( A and B ) = 0.037974684 0.0380

The probability that both households are dissatisfied with their purchase is 0.0380