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Given that the events A and B are mutually exclusive such that P(A)=12P(A)=\dfrac{1}{2}, P(A∪B)=23P(A\cup B)=\dfrac{2}{3} and P(B)=pP(B)=p, then the value of pp is:

Solution

✅ Correct Option: 1

Mutually exclusive events have P(A∩B)=0P(A\cap B)=0, so P(A∪B)=P(A)+P(B)P(A\cup B)=P(A)+P(B). Then p=23−12=16p=\dfrac23-\dfrac12=\dfrac16.

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