Question

Assertion :A bag contains $n+1$ coins. It is known that one of these coins has a head on both sides while the other coins are fair. One coins is selected at random and tossed. If head turns up, the probability that the selected coin was fair, is $\frac{n}{n+2}$ Reason: If an event $A$ occurs with two mutually exclusive and exhaustive events $E_1$ and $E_2$, then

$P\left(\frac{E_i}{A}\right)=\frac{P\left(E_i\right)P\left(\frac{A}{E_i}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)},i=1,2$

• 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 A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let $E=$ {selected coin is fair}

and $A=$ {head turns up}.

Thus, $P\left(A\right)=P\left(E\right)P\left(\frac{A}{E}\right)+P\left(\overline{E}\right)P\left(\frac{A}{\overline{E}}\right)$

$=\left(\frac{n}{n+1}\right)\frac{1}{2}+\left(\frac{1}{n+1}\right)1=\frac{n+2}{2(n+1)}$

Hence, $P\left(\frac{E}{A}\right)=\frac{P\left(E\right)P\left(\frac{A}{E}\right)}{P\left(A\right)}=\frac{\frac{n}{2(n+1)}}{\frac{n+2}{2(n+1)}}=\frac{n}{n+2}.$