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Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F)\) ? b. Suppose that for events \(A\) and \(B, P(A)=0.24, P(B)=0.24\), and \(P(A \cup B)=0.48 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Short Answer

Expert verified
i. The value of \(P(E \cap F)\) is 0. ii. The value of \(P(E \cup F)\) is 0.9. b. Yes, events A and B are mutually exclusive.

Step by step solution

01

Compute the Intersection of Mutually Exclusive Events

For mutually exclusive events, the intersection, represented by \(P(E \cap F)\), is always zero because mutually exclusive events cannot occur at the same time. Since E and F are mutually exclusive: \(P(E \cap F)=0\). This is always true and no computation is needed.
02

Compute the Union of Mutually Exclusive Events

For mutually exclusive events, the probability of their union is the sum of their individual probabilities. So, \(P(E \cup F)\) = \(P(E)\) + \(P(F) = 0.14 + 0.76 = 0.9\).
03

Check if Events A and B are Mutually Exclusive

Two events A and B are mutually exclusive if the probability of their union is equal to the sum of their individual probabilities. Therefore, if \(P(A \cup B) = P(A) + P(B)\), then A and B are mutually exclusive. Let's see if that is true. Given values are: \(P(A) = 0.24\), \(P(B) = 0.24\), and \(P(A \cup B) = 0.48\). From calculation, \(P(A) + P(B) = 0.24 + 0.24 = 0.48\). As this equals to \(P(A \cup B)\), we conclude A and B are mutually exclusive.

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