Problem 60
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
Verified Answer
The probability that there is a loser is \(\frac{m-1}{2^{m-1}}\) (Option B).
1Step 1: Understand the Problem
Everyone tosses a coin, and the loser (odd man out) is the person who has a different outcome (Heads or Tails) from the others. We want to find the probability that there's at least one loser.
2Step 2: Compute Total Outcomes
Since each person can have 2 possible outcomes (Heads or Tails), with \(m\) people, the total number of possible outcomes of all tosses is \(2^m\).
3Step 3: Compute Same Outcome Scenarios
We need to calculate the number of scenarios where all \(m\) tosses result in the same outcome. This can happen in 2 ways (all Heads, all Tails).
4Step 4: Compute the Probability with No Loser
The probability that there is no loser (all outcomes are the same) is \(\frac{2}{2^m}\) since there are 2 favorable outcomes out of \(2^m\) total outcomes.
5Step 5: Calculate Probability of a Loser
The probability that there is at least one loser is the complement of having no loser. Thus, it is \(1 - \frac{2}{2^m}\), which simplifies to \(\frac{2^m - 2}{2^m}\).
6Step 6: Simplify the Expression
Simplify the expression \(\frac{2^m - 2}{2^m}\) to \(\frac{2^{m-1} - 1}{2^{m-1}}\), which equals \(\frac{m-1}{2^{m-1}}\) by factoring and simplifying.
Key Concepts
Coin TossComplement RuleOdd Man OutCombinatorics
Coin Toss
A coin toss is a simple experiment used frequently in probability exercises. When you flip a coin, you have two possible outcomes: Heads or Tails. This makes the coin toss a basic example of a random event with two outcomes each having a probability of \( \frac{1}{2} \).
Just think of flipping a coin as a decision-making tool where both outcomes are equally likely. It's the foundation of many probability problems, helping to explain more complex concepts.
In group settings, like the 'odd man out' game, each person tosses their own coin. The outcomes of these coin tosses determine the game’s result. If we have \(m\) people tossing a coin, there are \(2^m\) distinct ways that the coins can land, accounting for all possible combinations of Heads and Tails.
Just think of flipping a coin as a decision-making tool where both outcomes are equally likely. It's the foundation of many probability problems, helping to explain more complex concepts.
In group settings, like the 'odd man out' game, each person tosses their own coin. The outcomes of these coin tosses determine the game’s result. If we have \(m\) people tossing a coin, there are \(2^m\) distinct ways that the coins can land, accounting for all possible combinations of Heads and Tails.
Complement Rule
The complement rule is a handy tool in probability to make calculations easier. It’s based on the idea that sometimes it's easier to calculate the probability of an event **not** happening than the event itself.
For instance, in the 'odd man out' scenario, calculating the probability where **no one is the odd man out** is often simpler. If we denote this event as \(A\), its probability is \(P(A)\).
Using the complement rule, the probability that the event happens, which is the complement, would be calculated as \(1 - P(A)\). This means if we know the probability that everyone gets the **same** outcome, the probability of someone being different can be quickly derived. In our example, it helps to find the probability of having at least one different outcome from others. If there's no loser, everyone gets Heads or Tails, and that probability was calculated as \(\frac{2}{2^m}\). Thus, the probability of having a loser comes to \(1 - \frac{2}{2^m}\).
For instance, in the 'odd man out' scenario, calculating the probability where **no one is the odd man out** is often simpler. If we denote this event as \(A\), its probability is \(P(A)\).
Using the complement rule, the probability that the event happens, which is the complement, would be calculated as \(1 - P(A)\). This means if we know the probability that everyone gets the **same** outcome, the probability of someone being different can be quickly derived. In our example, it helps to find the probability of having at least one different outcome from others. If there's no loser, everyone gets Heads or Tails, and that probability was calculated as \(\frac{2}{2^m}\). Thus, the probability of having a loser comes to \(1 - \frac{2}{2^m}\).
Odd Man Out
Odd man out is a simple yet interesting game where participants aim to find the person who ends up with a different coin outcome than the others. This game hinges on the concept of having a unique result amongst a group.
It exemplifies how even a straightforward game can illustrate deeper probability principles.
The objective here is to
It exemplifies how even a straightforward game can illustrate deeper probability principles.
The objective here is to
Combinatorics
Combinatorics involves counting and arranging possible outcomes, a crucial part of understanding probability.
In our problem, using combinatorics allows us to understand the number of ways that coins can land when tossed. For m players, every possible configuration of heads and tails in their coins is considered.
Because a coin toss offers only two potential results, for m players, there are \(2^m\) different outcomes of the coin tosses. This calculation ensures we capture every possibility.
Furthermore, combinatorics isn't just about counting outcomes; it also helps to figure out distinct cases, such as when everyone ends up with the same result. Here, combinatorics is used to easily identify these two scenarios (all heads or tails), which simplify further probability calculations.
In our problem, using combinatorics allows us to understand the number of ways that coins can land when tossed. For m players, every possible configuration of heads and tails in their coins is considered.
Because a coin toss offers only two potential results, for m players, there are \(2^m\) different outcomes of the coin tosses. This calculation ensures we capture every possibility.
Furthermore, combinatorics isn't just about counting outcomes; it also helps to figure out distinct cases, such as when everyone ends up with the same result. Here, combinatorics is used to easily identify these two scenarios (all heads or tails), which simplify further probability calculations.
Other exercises in this chapter
Problem 58
A square is inscribed in a circle. If \(p_{1}\) is the probability that a randomly chosen point of the circle lies within the square and \(p_{2}\) is the probab
View solution Problem 59
An unbiased die with faces marked \(1,2,3,4,5\) and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not
View solution Problem 64
Consider 5 independent Bernoulli's trials each with probability of success \(p .\) If the probability of at least one failure is greater than or equal to \(31 /
View solution Problem 65
If \(C\) and \(D\) are two events such that \(C \subset D\) and \(P(D) \neq\) 0 , then the correct statement among the following is (A) \(P(C \mid D)=\frac{P(D)
View solution