Question

Assertion :A fair die is thrown twice. Let (a, b) denote the outcome in which the first throw shows a and the second shows b. Let A and B be the following two events.
$A=\left\{\right(a,b\left)|\text{a is even}\right\},\left\{\right(a,b\left)|\text{b is even}\right\}$

If $C=\left\{\right(a,b\left)|\text{a+b is odd}\right\}$, then $P(A\cap B\cap C)=1/8$.

Reason: If $D=\left\{\right(a,b\left)|\text{a+b is even}\right\}$, then $P(A\cap B\cap D|A\cup B)=1/3$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

Assertion $\xi$ Reason :

$A=$ The no. appearing on the dice when it is thrown twice

$C=$ The sum of no. appearing be $(a+b)$

Let the sum of two outcomes be $(a+b)$

Given $A$ $\xi$ $B$ be the following two events

$\Rightarrow D=\left[\right(a,b)|a+biseven]$, then $P(A\cap B\cap D|A\cup B)$

$\Rightarrow P(A\cap B\cap D)=P(A\cup B\cup D)-P\left(A\right)+P\left(B\right)+P\left(D\right)-P(A\cap B)-P(B\cap D)-P(A\cap D)$

$\Rightarrow P(A\cup B)=P\left(A\right)+P\left(B\right)$

$\Rightarrow P(A\cup B\cup D)=P\left(A\right)+P\left(B\right)+P\left(D\right)$

$\Rightarrow P(A\cap B\cap D)=P(A\cup B\cup D)-P(A\cup B\cup C)+P(A\cap B)+P(B\cap D)+P(A\cap D)$

$\therefore P(A\cap B\cap D)=P(A\cap B)-P(B\cap C)+P(A\cap C)$

$\Rightarrow P(A\cap B\cap D|A\cup B)=P(B\cap C)+P(A\cap C)$

$=\frac{2}{6}=\frac{1}{3}.$

Hence, the answer is Assertion is incorrect but reason is correct.