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. Can't say

Answer 1

The correct option is A $\frac{1}{2}$

Let ${\lambda }^{\prime },{\mu }^{\prime }$ be the numbers of heads and tails thrown by A and B respectively , so that $\lambda {\lambda }^{\prime }=n+1$ $\mu +{\mu }^{\prime }=n.$
The required probability P is the probability of the inequality $\lambda >\mu .$
The probability 1-p of the opposite event $\lambda \le \mu$ is at the same time the probability of the inequality ${\lambda }^{\prime }>{\mu }^{\prime }.$ that is, 1-P is the probability that A will throw more tails than B. [Reason :$\lambda \le \mu \Rightarrow n+1-{\lambda }^{\prime }\le n-{\mu }^{\prime }$
$\Rightarrow 1-{\lambda }^{\prime }\le -{\mu }^{\prime }\Rightarrow {\lambda }^{\prime }-1\ge {\mu }^{\prime }$
$\Rightarrow {\lambda }^{\prime }\ge {\mu }^{\prime }+1>{\mu }^{\prime }]$ By reason of symmetry, $1-P=P$ or $P=\frac{1}{2}.$