The probabilities of happening of the events A and B are $\frac{1}{4}$ and $\frac{1}{2}$ respectively. If the probability of happening of A and B simultaneously is $\frac{7}{50}$, then probability of neither A nor B happening, is equal to
Answer & explanation
Correct answer: option 1
$P(A)=\frac{1}{4}, P(B)=\frac{1}{2}, P(A \cap B)=\frac{7}{50}$
Now, $P(\bar{A} \cap \bar{B})=P(\bar{A})+P(\bar{B})-P(\bar{A} \cup \bar{B})$
$=1-P(A)+1-P(B)-(1-P(A \cap B))$
$=1-\frac{1}{4}-\frac{1}{2}+\frac{7}{50}=\frac{39}{100}$