Answer to Question #139506 in Statistics and Probability for ReeC

Question

Supposed you just received a shipment of fourteen 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

There are 3 defective TV and 11 working TV.

The probability that we select two TVs that do work are

$\displaystyle P(A) = \frac{11}{14} \cdot \frac{10}{13} = \frac{55}{91}$

The condition "at least one of the two televisions does not work" will be satisfied if 1st - defective, 2nd - working, or 1st - working, 2nd - defective, or both are defective.

The probability of this is

$\displaystyle P(B) = \frac{3}{14} \cdot \frac{11}{13} + \frac{11}{14} \cdot\frac{3}{13} + \frac{3}{14} \cdot\frac{2}{13} = \frac{72}{182} = \frac{36}{91}$

Answer: 55/91 and 36/91.