Problem 60 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In a game "odd man out" each of \(m \geq 2\) persons, tosses a coin to determine who will buy refreshments for the entire group. The odd man out is the one with a different outcome from the rest. The probability that there is a loser in any game is (A) \(\frac{1}{2^{m-1}}\) (B) \(\frac{m-1}{2^{m-1}}\) (C) \(\frac{m}{2^{m-1}}\) (D) none of these

Step-by-Step Solution

Answer

Option (C) \(\frac{m}{2^{m-1}}\) is correct.

Step 1: Understand the Game

In the game "odd man out", each player tosses a coin. A loser is defined as the player with a different coin outcome from the rest. For example, if everyone but one person has heads, the odd one with tails is the loser.

Step 2: Define the Outcomes

When each player tosses a coin, there are two possible outcomes: heads (H) or tails (T). For any collection of coin tosses, if one outcome is unique while the others are identical, there is a loser.

Step 3: Calculate Total Possible Outcomes

Each of the \(m\) players can have two possible outcomes. Therefore, the total number of outcome combinations is \(2^m\) (since each player tosses their coin once).

Step 4: Calculate Outlier Patterns

For there to be a loser, exactly one person must have a different outcome from the other \(m-1\). This can occur in two ways: one player is heads when the others are tails, or one player is tails when the others are heads.

Step 5: Count Favorable Outlier Outcomes

If one person is the odd man out with heads and all others are tails, or vice versa, these are the two possible outlier situations. Each outlier scenario can occur in \(m\) different ways (since any of the \(m\) players could be the loser). Thus, there are \(2m\) favorable outcomes.

Step 6: Calculate the Probability

The probability of having a loser is equal to the number of favorable outcomes divided by the total number of outcomes: \(\frac{2m}{2^m} = \frac{m}{2^{m-1}}\).

Step 7: Match to Options

Compare the calculated probability \(\frac{m}{2^{m-1}}\) with the provided options. Option (C) matches our solution.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and finding patterns in sets. It is especially important in probability, where we count all possible outcomes. In the context of the game "odd man out," combinatorics is used to understand how many different ways the players can toss their coins.

To break it down, if you have a situation with multiple participants, like in this coin toss game, combinatorics helps us calculate all possible results of their actions. So, if there are \(m\) players, each having two potential results for their coin toss—heads or tails—we need to count all these possibilities. This is where we use the concept of permutations and combinations in combinatorics to simplify our calculations.

Understanding basic combinatorics sets the stage for calculating the probability of an event by determining the number of desired outcomes and total outcomes.

Coin Toss

A coin toss is a simple random experiment where a coin is flipped, resulting in one of two possible outcomes: heads (H) or tails (T). In probability, each outcome is considered equally likely if the coin is fair.

In the game scenario, each player independently tosses a coin, resulting in either a head or a tail. Given there are \(m\) players, and each has two outcomes, combinatorics tell us that the total possible outcomes for the group are \(2^m\).

Coin toss outcomes, when considered together, help us make predictions about the probability of different events. This simple yet foundational concept in probability helps illustrate how individual events can combine to form a wide range of possible scenarios.

Permutations

Permutations in probability refer to the arrangement or sequence of outcomes. In the "odd man out" game, permutations help us consider all possible scenarios where one player differs from the others. When we want to find out which player will be the odd one with a different toss, we consider permutations.

For instance, one player having a head while all others have tails is a situation determined by permutations. Each player can uniquely be the odd one, so we have \(m\) different permutations here. This repeats for the situation where one player has tails and others have heads, leading to a total of \(2m\) favorable outcomes, considering both permutations and combinations.

Using permutations ensures we count all possible distinct arrangements leading to the specific desired outcome, crucial for accurate probability calculation.