Question

Two persons $A$ and $B$ have respectively $n+1$ and $n$ coins, which they toss simultaneously. Then the probability that $A$ will have more heads than $B$ is:

• A. $\frac{1}{2}$
• B. $>\frac{1}{2}$
• C. $<\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option:

Let $a$ and $a\text{'}$ be the number of heads and tails by $A$.

So, $a+a\text{'}=n+1$ $...\left(1\right)$

Let $b$ and $b\text{'}$ be the number of heads and tails by $B$.

So, $b+b\text{'}=n$

Required probability is that $A$ has more heads than $B$, so let, $P(a>b)=P(a\text{'}>b\text{'})=p$

$\Rightarrow P(a>b)=1-P(a\le b)$

Now for, $a\le b$

$$\begin{gathered} \Rightarrow n+1-a\text{'}\le n-b\text{'} \\ \Rightarrow 1+b\text{'}\le a\text{'} \end{gathered}$$

Thus, $b\text{'}<a\text{'}$, so we can say that $P(a\le b)=P(b\text{'}<a\text{'})$

$$\begin{gathered} \Rightarrow P(a>b)=1-P(b\text{'}<a\text{'}) \\ \Rightarrow p=1-p \\ \Rightarrow 2p=1 \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$

Therefore, option (A) is the correct answer.