Question

Let $A$ and $B$ be two events such that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$. If $0<P\left(A\right)<1$ and $0<P\left(B\right)<1$, then $P(A\cup B{)}^{\prime }$ is equal to

• A. $1-P\left(A\right)$
• B. $1-P\left(A^{\prime }\right)$
• C. $1-P\left(A\right)P\left(B\right)$
• D. $[1-P(A\left)\right]P\left(B^{\prime }\right)$
• E. $1$

Answer 1

Answer: (D)

$[1-P(A\left)\right]P\left(B^{\prime }\right)$

Given, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$
Therefore, $P(A\cup B{)}^{\prime }=1-P(A\cup B)$
$=1-\left[P\right(A)+P(B)-P(A\left)P\right(B\left)\right]$
$=1[1-P(A\left)\right]-P\left(B\right)[1-P(A\left)\right]$
$=[1-P(A\left)\right][1-P(B\left)\right]$