Question

Suppose you just received a shipment of twelve 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? The probability that both televisions work is nothing

Answer 1

Answer: The probability that both televisions work is 0.5625 .

The probability at least one of the two televisions does not​ work is 0.4375.

Step-by-step explanation:

Given : Number of televisions received in shipment= 12

Number of defective televisions received in shipment= 3

Then proportion of defective televisions =$\dfrac{3}{12}=0.25$

Using binomial probability formula, the probability of getting success in x trials is given by :-

$P(x)=^nC_xp^x(1-p)^{n-x}$

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

$P(0)=^2C_0(0.25)^0(0.75)^2=(0.75)^2=0.5625$

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

$P(x\geq1)=1-P(0)=1-0.5625=0.4375$