First slide

Introduction to probability

Question

In a game called "odd man out" m(m > 2) persons toss a coin to determine who will buy refreshments for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is

Moderate

A

$\frac{1}{2 m}$
B

$\frac{m}{2^{m - 1}}$
C

$\frac{2}{m}$
D

none of these

Solution

Let A denote the event that there is an odd man out in a game. The total number of possible cases is 2^m. A person is odd man out if he is alone in getting a head or a tail.
The number of ways in which there is exactly one tail (head) and the rest are heads (tails) is $^{m} C_{1} = m$ . Thus, the number of favorable ways is m + m = 2m. Therefore,
$P (A) = \frac{2 m}{2^{m}} = \frac{m}{2^{m - 1}}$

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