Probability $P(A) = \frac{4}{5},P\left( B^{'} \right) = \frac{2}{5}$and $P(A \cap B) = \frac{1}{2},$then$P(A \cap B')$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Here, $P(B) = 1 - P\left( B^{'} ight) = 1 - \frac{2}{5} = \frac{3}{5}$
$	herefore\ P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= \frac{4}{5} + \frac{3}{5} - \frac{1}{2} = \frac{9}{10}$
$	herefore\ P\left( A \cap B^{'} ight) = P(A \cup B) - P(B)$
$= \frac{9}{10} - \frac{3}{5} = \frac{3}{10}$

Q. Probability $P(A) = \frac{4}{5},P\left( B^{'} \right) = \frac{2}{5}$and $P(A \cap B) = \frac{1}{2},$then$P(A \cap B')$ is equal to

Explanation

Reviewed by Dr. Abubacker

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