Question

The probability of simultaneous occurrence of 2 events A & B is $p$. If the probability that exactly one of A, B occurs is $q$, then which of the following alternatives is INCORRECT?

• A. $P\left(\overline{A}\right)+P\left(\overline{B}\right)=2+2q-p$
• B. $P\left(\overline{A}\right)+P\left(\overline{B}\right)=2+2p-q$
• C. $P\left(\right(A\cap B\left)/\right(A\cup B\left)\right)=\frac{P}{p+q}$
• D. $P(\overline{A}\cap \overline{B})=21-p-q$

Answer 1

The correct option is A $P\left(\overline{A}\right)+P\left(\overline{B}\right)=2+2q-p$

$P(A\cap B)=p$ and $P\left(A\right)+P\left(B\right)-2P(A\cap B)=q$
$\Rightarrow P\left(A\right)+P\left(B\right)-2p=q$
$\Rightarrow P\left(A\right)+P\left(B\right)=2p-q$
$\Rightarrow 1-P\left(\overline{A}\right)+1-P\left(\overline{B}\right)=2p+q$
$\Rightarrow P\left(\overline{A}\right)+P\left(\overline{B}\right)=2-2p-q$
So, alternative (b) is correct.

Now,
$P\left\{\right(A\cap B\left)/\right(A\cup B\left)\right\}=\frac{P\left[\right(A\cap B)\cap (A\cup B\left)\right]}{P(A\cup B)}$
$\Rightarrow P\left\{\right(A\cap B\left)/\right(A\cup B\left)\right\}=\frac{P(A\cap B)}{P(A\cup B)}$
$\Rightarrow P\left\{\right(A\cap B\left)/\right(A\cup B\left)\right\}=\frac{P\left[\right(A\cap B)}{P\left(A\right)+P\left(B\right)-P(A\cap B)}$
$\Rightarrow P\left\{\right(A\cap B\left)/\right(A\cup B\left)\right\}=\frac{P}{2p+q-p}=\frac{p}{p+q}$
So, alternative (c) is correct.
Finally,

$P(\overline{A}\cap \overline{B})=P\left(\overline{A\cup B}\right)=1-P(A\cup B)$
$\Rightarrow P(\overline{A}\cap \overline{B})=1-\left[P\right(A)+P(B)-P(A\cap B\left)\right]$
$\Rightarrow P(\overline{A}\cap \overline{B})=1-[2p+q-p]=1-p-q$
So, alternative (d) is correct.
Hence, alternative (a) is incorrect.