First slide

Theorems of 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

$1 / 2 m$
B

$m / 2^{m - 1}$
C

$2 / m$
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 case s 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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