# The probability of the event.

The events $A$, $B$, and $C$ occur with respective probabilities $0.26$ , $0.25$, and $0.41$. The events $B$ and $C$ are mutually exclusive; likewise the events $B$ and $A$ are mutually exclusive. The probability of the event $C\cap A$ is 0.02. Compute the probability of the event $C\cup (A \cap B')$.

I tried to solve it: $$P(C\cup(A\cap B')){= P((C\cup A) \cap (C\cup B')) \\=(P(C)+P(A)-P(C\cap A)) \cdot (P(C)+P(B')-P(C\cap B'))\\=(0.41 - 0.26 - 0.02)\cdot(0.41 + 0.74 - 0.3034)\\=0.65\cdot 0.8466\\=0.55029\\=}$$

I am not sure whether it is right or wrong. If it is wrong, then please tell me the correct answer. It would be much appreciated!

## 1 Answer

Hint: Since $A$ and $B$ are mutually exclusive, $A$ is completely contained in $B'$. In other words, $A \cap B' = A$.

So you can use the values of $P(A), P(C), P(C \cap A)$ to get the answer.

As to your answer, there's a problem with the 2nd step because effectively, you're writing $P(X \cap Y) = P(X) \cap P(Y)$, which makes no sense since $P(X)$ and $P(Y)$ are numbers, not sets.

Edit: $P(C \cup (A\cap B')) = P(C\cup A) = P(C) + P(A) -P(C \cap A) = 0.26 + 0.41 - 0.02 = 0.65$.

• so is it like =0.41+0.26-0.02*0.41+0.74-0.41=0.481 ? Is it like this? – johnc Apr 6 '18 at 6:42
• How did you get that expression? – Shirish Kulhari Apr 6 '18 at 6:43
• You mention that A∩B′=A . so here =(P(C)+P(A)-P(C∩A)) ∩ (P(C)+P(B')-P(C∩B')) I changed P(C∩B') to P(C) – johnc Apr 6 '18 at 6:45
• Could you solve this problem for me? could you show me the steps as well? It would be easier to understand. – johnc Apr 6 '18 at 6:47
• I'll edit my answer to show the steps. One thing though - you're probably mistaking $P(C \cap A)$ as $P(C)*P(A)$. This is true only if $C$ and $A$ are independent events, which isn't mentioned. You need to review the basics of both set theory and probability. "Introduction to Probability" by Bertsekas and Tsitsiklis is a good reference. – Shirish Kulhari Apr 6 '18 at 7:00