Question

Assume that A and B are events in a probability space with the property that P(A) = 0.5, P(B) = 0.6, and P(A ∪ B) = 0.9.

1. Explain why A and B cannot be independent.

2. Is A favorable or unfavorable to B? (Remember that an event E is said to be favorable to F if P(F|E) > P(F); that is, if the knowledge that E occurred increases the plausibility of F.)

Answer 1

Given and $P\left ( A\cup B \right )=0.9$ .

Now

$P\left ( A\cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )\\ 0.9=0.5+0.6-P\left ( A\cap B \right )\\ P\left ( A\cap B \right )=0.2$

a) If A and B are independent, then $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$ . Here

$P\left ( A\cap B \right )=0.2\neq 0.3=0.5\times 0.6=P\left ( A \right )P\left ( B \right )$

Hence A and B cannot be independent.

b) We have using conditional probability $P\left ( B|A \right )=P\left ( A\cap B \right )/P\left ( A \right )=0.2/0.5=0.4$

Since $P\left ( B|A \right )=0.4<0.6=P\left ( B \right )$ , A is unfavorable to B.