Question

D) If P(A) = 0.6, P(B) = 0.5, P(AB) = 0.3 then find i) P(A) ii) P(AUB) iii) P(A'B) iv) P(A' B')​

Answer 1

The student has been helped with finding the probability of A, A or B, not A and B, and neither A nor B, using given probabilities and probability rules.so P(A' B') = 1 - 0.8 = 0.2.

Step-by-step explanation:

The student is asking for help with finding the probabilities of different combinations and conditions of two events A and B. Given that P(A) = 0.6, P(B) = 0.5, and P(AB) = 0.3 (the probability that both A and B occur), we are to find the following:

1. P(A): This is given directly as 0.6.
2. P(AUB): The probability of A or B or both. Using the formula P(AUB) = P(A) + P(B) - P(AB), we get P(AUB) = 0.6 + 0.5 - 0.3 = 0.8.
3. P(A'B): The probability of not A and B. Using the complement rule, P(A') = 1 - P(A), so P(A') = 0.4. Therefore, P(A'B) = P(A')P(B), given that A and B are not necessarily independent, we cannot assume P(A'B) = P(A')P(B). Instead, we use P(A'B) = P(B) - P(AB) = 0.5 - 0.3 = 0.2.
4. P(A' B'): The probability of not A and not B. This is found by P(A' B') = 1 - P(AUB), so P(A' B') = 1 - 0.8 = 0.2.