Answer:
0.875
Step-by-step explanation:
Let the probability that Billy passes test 1 be P(A)
Let the probability that Billy passes test 2 be P(B)
P(A n B) = 0.7
P(A) = P(B)
Let them both be numerically x.
P(A) = P(B) = x
P(A') = P(B') = (1 - x)
Probability that he passes at least one test = 0.9
But (probability that he passes at least 1 test) = 1 - (Probabilty that he fails both tests)
Probability that he fails both tests = 1 - 0.9 = 0.1
P(A' n B')= 0.1
P(U) = P(A) - P(A n B) + P(B) - P(A n B) + P(A n B) + P(A' n B') = 1
x - 0.7 + x - 0.7 + 0.7 + 0.1 = 1
2x = 1 + 0.7 - 0.1 = 1.6
x = 0.8
P(A) = P(B) = 0.8
The conditional probability of Billy passing test 2 given the event that he passes test 1 = P(B|A) = P(A n B)/P(A) = (0.7/0.8)
P(B|A) = 0.875
Hope this helps!!!