JEE Main 2022 — Mathematics Probability & Statistics

Given,

$A$A and $B$B be two events such that $P(B\mid A)=\frac{2}{5}$P(B∣A)=52​, $P(A\mid B)=\frac{1}{7}$P(A∣B)=71​ and $P(A\cap B)=\frac{1}{9}$P(A∩B)=91​,

Now using the formula of conditional probability we get,

$$P(A\mid B)=\frac{1}{7}\Rightarrow \frac{P(A\cap B)}{P(B)}=\frac{1}{7}$$P(A∣B)=71​⇒P(B)P(A∩B)​=71​

$$\Rightarrow \frac{\frac{1}{9}}{P(B)}=\frac{1}{7}$$⇒P(B)91​​=71​

$$\Rightarrow P(B)=\frac{7}{9}$$⇒P(B)=97​

Now again using formula we get,

$$P(B\mid A)=\frac{2}{5}\Rightarrow \frac{P(A\cap B)}{P(A)}=\frac{2}{5}$$P(B∣A)=52​⇒P(A)P(A∩B)​=52​

$$\Rightarrow \frac{\frac{1}{9}}{P(A)}=\frac{2}{5}$$⇒P(A)91​​=52​

$$\Rightarrow P(A)=\frac{5}{18}$$⇒P(A)=185​

Now we know that,

$$P({A}^{'}\cup B)=1-P(A\cup B)+P(B)$$P(A′∪B)=1−P(A∪B)+P(B)

$$\Rightarrow P({A}^{'}\cup B)=1-P(A)+P(A\cap B)=\frac{5}{6}$$⇒P(A′∪B)=1−P(A)+P(A∩B)=65​

And $P({A}^{'}\cap {B}^{'})=1-P(A\cup B)$P(A′∩B′)=1−P(A∪B)

{also $P(A\cup B)=P(A)+P(B)-P(A\cap B)$P(A∪B)=P(A)+P(B)−P(A∩B)}

$$\Rightarrow P({A}^{'}\cap {B}^{'})=1-P(A)-P(B)+P(A\cap B)=\frac{1}{18}$$⇒P(A′∩B′)=1−P(A)−P(B)+P(A∩B)=181​

$\Rightarrow$⇒ Both $(S1)$(S1) and $(S2)$(S2) are true.