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

Infinity Learn · Accepted Answer

Let A denote the event that there is an odd man out in a game. The total number of possible cases is $$2m$$. 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 $$mC1=m$$. Thus, the number of favorable ways is m $$+$$ m $$=$$ $$2m$$. Therefore,$$P(A)=2m2m=m2m$$−$$1$$

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

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