# 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

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

(A) 1/2m

(B) $\dfrac{m}{2^{m-1}}$

(C) 2/m

(D) none of these

Answer : option (B)

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 $mC_{1}=m$.

Thus, the number of favourable ways is m + m = 2m.

Therefore, $P\left( A\right) =\dfrac{2m}{2^{m}}=\dfrac{m}{2^{m-1}}$

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