Final answer:
a) The probability that the third TV is good, given that the first two were good, is 0.8. b) The probability that the third TV is defective, given that one of the other two was good and the other defective, is 0.0909.
Step-by-step explanation:
To calculate the probability of these events, we need to use the concept of conditional probability. Let's break down each event:
a) The probability that the third TV is good, given that the first two TVs were good, can be calculated as:
P(G3 | G1 and G2) = P(G3 and G1 and G2) / P(G1 and G2)
Since the TVs are chosen one by one, the probability that the third TV is good is the same as the probability that the first TV is good:
P(G3 | G1 and G2) = P(G) = 16/20 = 0.8
b) The probability that the third TV is defective, given that one of the other two TVs was good and the other defective, can be calculated as:
P(D3 | (G1 and D2) or (D1 and G2)) = P(D3 and ((G1 and D2) or (D1 and G2))) / P(((G1 and D2) or (D1 and G2)))
There are two possibilities: 1. G1 and D2 (first TV is good and second TV is defective) and 2. D1 and G2 (first TV is defective and second TV is good). Since these events are mutually exclusive, we can calculate the probability as:
P(D3 | (G1 and D2) or (D1 and G2)) = (P(D3 and (G1 and D2))) / (P(G1 and D2) + P(D1 and G2))
P(D3 | (G1 and D2) or (D1 and G2)) = (P(D3 and G1 and D2)) / (P(G1 and D2) + P(D1 and G2))
P(D3 | (G1 and D2) or (D1 and G2)) = (P(D3) * P(G1) * P(D2)) / ((P(G1) * P(D2)) + (P(D1) * P(G2)))
Substituting the given probabilities, we get:
P(D3 | (G1 and D2) or (D1 and G2)) = (4/20 * 16/20 * 4/19) / ((16/20 * 4/20) + (4/20 * 16/20)) = 0.0909