Question

Suppose you just received a shipment of eleven televisions. Three of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?

Answer 1

Answer: The probability that both televisions work : 0.5329

The probability at least one of the two televisions does not​ work : 0.4671

Explanation:

Given : The total number of television : 11

The number of defective television : 3

The probability that the television is defective :
`p=(3)/(11)\approx0.27`

Binomial distribution formula :-

`P(x)=^nC_xp^x(1-p)^(n-x)`, where P(X) is the probability of getting success in x trials, p is the probability of success and n is the total trials.

If two televisions are randomly​ selected, then the probability that both televisions work:

`P(0)=^2C_0(0.27)^0(1-0.27)^(2-0)=(1)(0.73)^2=0.5329`

The probability at least one of the two televisions does not​ work :

`P(X\geq1)=1-P(0)=1-0.5329=0.4671`