Question

Assertion :A throws $n+1$ coins and $B$ throws $n$ coins. The chance that $A$ has greater number of heads than $B$ is $\frac{1}{2}$ Reason: $\sum _{0<k<m\le n}^{}\sum _{}^{}{}^{n+1}{C}_m{.}^n{C}_k=\frac{1}{2}.2^{2n+1}$

• 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 $P(m,k)$ be the probability that $A$ throws $m$ heads and $B$ throws $k$ heads.

Then,$P(m,k){=}^{n+1}{C}_m.{\left(\frac{1}{2}\right)}^m{\left(\frac{1}{2}\right)}^{n+1-m}{.}^n{C}_k.{\left(\frac{1}{2}\right)}^k.{\left(\frac{1}{2}\right)}^{n-k}$

${=}^{n+1}{C}_m{.}^n{C}_k.{\left(\frac{1}{2}\right)}^{2n+1}$

and, required probability $\sum _{0<k<}^{}\sum _{m\le n+1}^{}\frac{{}^{n+1}{C}_m{.}^n{C}_k}{2^{2n+1}}$

Now, $\sum _{0<k<}^{}\sum _{m\le n}^{}{}^{n+1}{C}_m{.}^n{C}_k$

$=\frac{1}{2}({}^{n+1}{C}_0{+}^{n+1}{C_1+}^{n+1}{C}_2{+}^{n+1}{C}_3+...{+}^{n+1}{C}_{n+1})$

$\times ({}^n{C_0+}^n{C_1+}^n{C}_2{+}^n{C}_3+...{+}^n{C}_n)$

$=\frac{1}{2}.2^{n+1}{2}^n=\frac{1}{2}.2^{n+1}$

Thus, the required probability $=\frac{1}{2}$