1. Probability

A and B are two independent events in the sampling space S. Given that P(A U B) = 5/8 and P(A ∩ B')= 7/24, Find

(a) P(B)
(b) P(A ∩ B)
(c) P(A)
(d) P(A' U B')

The answer for a,b,c,d are 1/3, 7/48, 7/16, 41/48 respectively.....I'm not sure how to get those probabilities....

2. Do you know what independent events mean? Look at the definition of Independent Events and then show your attempt at the problem.

and $P(A \cup B) = P(A)+P(B)-P(A\cap B)$

3. Originally Posted by nav92
A and B are two independent events in the sampling space S. Given that P(A U B) = 5/8 and P(A ∩ B')= 7/24, Find

(a) P(B)
(b) P(A ∩ B)
(c) P(A)
(d) P(A' U B')

The answer for a,b,c,d are 1/3, 7/48, 7/16, 41/48 respectively.....I'm not sure how to get those probabilities....
$\Pr(A \cup B) = 1 - \Pr(A' \cap B') = \frac{5}{8} \Rightarrow \Pr(A' \cap B') = \frac{3}{8}$.

Let $\Pr(A \cap B) = x$.

Now try to draw a Karnaugh Table.

Note: $\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)$ since A and B are independent and so $x = \left(x + \frac{7}{24}\right) \cdot \frac{8}{24} \Rightarrow x = ....$

4. $\frac{5}{8} = P(A) + P(B) - P(A)P(B) = P(B) + P(A)P(B') = P(B)+\frac{7}{{24}}$

Solve for $P(B)$