Question

Assertion :A number is chosen at random from the numbers 1, 2, 3 ..... 6n. Let A & B be defined as follows
'A' : Number is multiple of 2
'B' : Number is multiple of 3. Reason: Two events A & B are independent then $P(A\cap B)=P\left(A\right)P\left(B\right)$

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
• B. Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion
• C. Assertion is true but Reason is false
• D. Assertion is false but Reason is true

Answer 1

The correct option is D Assertion is false but Reason is true

$A=$ multiple of2

$A=\{2,4,\mathrm{6....6}n\}$

$n\left(A\right)=3n$

$\therefore P\left(A\right)=\frac{3n}{6n}=\frac{1}{2}andB=\{3,6,9,\mathrm{12...},6n\}$

$\therefore A\cap B=\{6,12,18,...,6n-6\}$

$\Rightarrow n\left(B\right)=2n$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{2n}{6n}=\frac{1}{3}$

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3},soP\left(A\right)P\left(B\right)=\frac{1}{6}$

Now $n(A\cap B)=?$ (say $N$)

$6n-6=6+(N-1)6(\because t_n=a+(n-d))$

$\therefore N=n-1$

$\therefore P(A\cap B)=\frac{n-1}{6n}$

$\ne \frac{1}{6}=\frac{1}{2}\cdot \frac{1}{3}=P\left(A\right)P\left(B\right)$

$\therefore P(A\cap B)\ne P\left(A\right)P\left(B\right)$

$\Rightarrow$ Assertion (A) is false but Reason (R) is true.