# Probability of three independent events

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.

• 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. Oct 4, 2014 at 4:27
• 1. Draw a diagram. $\:$ 2. Apply basic probability rules Oct 4, 2014 at 9:52
• Also asked at math.stackexchange.com/q/957511/18398 Oct 8, 2014 at 3:59

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 $67.18\%$.