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Let the three independent events $ A, B,$ and $ C$ be such that $P(A)=P(B)=P(C)= \frac14.$ find $P[(A^*\cap B^*) \cup C].$

My solution starts from using the probability of their complements which is $\frac34$, I do not know how to answer this question. Please help.

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  • $\begingroup$ Hint: $C = (A\cap B\cap C)\cup(A*\cap B\cap C)\cup(A\cap B*\cap C)\cup(A*\cap B*\cap C)$ where the four events on the right are mutually exclusive. $\endgroup$
    – Dilip Sarwate
    Commented Oct 4, 2014 at 4:27
  • $\begingroup$ 1. Draw a diagram. $\:$ 2. Apply basic probability rules $\endgroup$
    – Glen_b
    Commented Oct 4, 2014 at 9:52
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    $\begingroup$ Also asked at math.stackexchange.com/q/957511/18398 $\endgroup$
    – JRN
    Commented Oct 8, 2014 at 3:59

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Remember that for any two events $U$ and $V$, the sum rule says $$ P(U\cup V)=P(U)+P(V)-P(U\cap V) \, . $$

Also, if $U$ and $V$ are independent, then $$ P(U^c\cap V^c)=1-P(U\cup V)=1-P(U)-P(V)+P(U\cap V)=(1-P(U))(1-P(V))=P(U^c)P(V^c) \, , $$ meaning that $U^c$ and $V^c$ are also independent.

This is all that you need. Just take $U=A^c\cap B^c$ and $V=C$, and do the algebra starting with the sum rule.

The answer is approximately $67.18\%$.

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