$A$ and $B$ are two independent events such that $P(A)\frac{1}{2}$ and $P(B) = \frac{1}{3},\ then\ P$(neither $A\$nor $B$)is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Since, $A$ and $B$ are independent events.
$	herefore\ P(A \cap B) = P(A).P(B) = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$
Now, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3}$
$	herefore\ P\left( \overline{A} \cap \overline{B} ight) = 1 - P(A \cup B)$
$= 1 - \frac{2}{3} = \frac{1}{3}$

Q. $A$ and $B$ are two independent events such that $P(A)\frac{1}{2}$ and $P(B) = \frac{1}{3},\ then\ P$(neither $A\$nor $B$)is equal to

Explanation

Reviewed by Dr. Abubacker

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