The probability of the simultaneous occurrence of two events A and B is p. If the probability that exactly one of A, B occurs is q, then which of the following alternatives is incorrect ? |
$P(\overline{A}+P\overline{B}) = 2 + 2q -p$ $P(\overline{A}+P\overline{B}) = 2 - 2q -p$ $P(A ∩ B/A ∪ B) = \frac{p}{p+q}$ $P(\overline{A} ∩ \overline{B}) = 1- p- q$ |
$P(\overline{A}+P\overline{B}) = 2 + 2q -p$ |
We have, $P(A∩ B)= p $ and $P(A) +P(B)-2P(A ∩ B)= q$ $⇒ P(A) + P(B) -2p = q $ $⇒ P(A) + P(B) =2p + q $ $⇒ 1-P(\overline{A}) +1 - P(\overline{B})= 2p+q$ $⇒P(\overline{A})+P(\overline{B})= 2 - 2p - q$ So, alternative (b) is correct. Now, $\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{P[(A ∩B)∩(A ∪B)}{P(A ∪B)}$ $⇒ P\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{P(A ∩B)}{P(A ∪B)}$ $⇒ P\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{P(A ∩B)}{P(A)+P(B)-P(A ∩B)}$ $⇒ P\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{p}{2p+q-p}=\frac{p}{p+q}$ So, alternative (c) is correct. Finally, $P(\overline{A} ∩\overline{B}) P(\overline{A ∪B})=1- P(A ∪B)$ $⇒P(\overline{A} ∩\overline{B})= P(\overline{A ∪B})=1 -[P(A) +P(B) -P(A ∩B)]$ $⇒P(\overline{A} ∩\overline{B})= P(\overline{A ∪B})=1-[2p+q-p]=1-p-q$ |