The probability that at least one of $A\ and\ B$ occurs is 0.6. If $A\ and\ B$ occur simultaneously with probability 0.3, then $P\left( A^{'} \right) + P(B^{'})$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Given $P(A \cup B) = 0.6,\ P(A \cap B) = 0.3$
$	herefore\ P\left( A^{'} ight) + P(B^{'})$
$= 1 - P(A) + 1 - P(B) = 2 - \left\{ P(A) + P(B) ight\}$
$= 2 - \left\{ P(A \cup B) + P(A \cap B) ight\}$
$= 2 - \left\{ 0.6 + 0.3 ight\} = 2 - 0.9 = 1.1$

Q. The probability that at least one of $A\ and\ B$ occurs is 0.6. If $A\ and\ B$ occur simultaneously with probability 0.3, then $P\left( A^{'} \right) + P(B^{'})$ is

Explanation

Reviewed by Dr. Abubacker

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