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Magazine Chapter 5: Problem 6
When Susan and Jessica play a card game, Susan wins \(60 \%\) of the time. If they play 9 games, what is the probability that Jessica will have won more games than Susan?
Short Answer
Expert verified
The probability that Jessica wins more than Susan in 9 games is approximately 0.382.
Step by step solution
01
Understand the problem
We need to determine the probability that Jessica wins more games than Susan in 9 games. This means we are looking for the probability that Jessica wins at least 5 games out of 9.
02
Establish key probabilities
We are given that Susan wins 60% of the games, so the probability of Susan winning any single game is \( P(S) = 0.6 \). Consequently, the probability that Jessica wins a single game is \( P(J) = 1 - P(S) = 0.4 \).
03
Define the random variable
Let \( X \) be a random variable representing the number of games Jessica wins. \( X \) follows a Binomial distribution with \( n = 9 \) (the number of games) and \( p = 0.4 \) (probability of Jessica winning a game).
04
Calculate probabilities of Jessica winning at least 5 games
We need the probability that \( X \geq 5 \). Calculate \( P(X = k) \) for \( k = 5, 6, 7, 8, \) and \( 9 \). The formula for binomial probability is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
05
Calculate probabilities for each scenario
Calculate each probability: - \( P(X = 5) = \binom{9}{5} (0.4)^5 (0.6)^4 \) - \( P(X = 6) = \binom{9}{6} (0.4)^6 (0.6)^3 \) - \( P(X = 7) = \binom{9}{7} (0.4)^7 (0.6)^2 \) - \( P(X = 8) = \binom{9}{8} (0.4)^8 (0.6)^1 \) - \( P(X = 9) = \binom{9}{9} (0.4)^9 (0.6)^0 \)
06
Sum the probabilities
Sum all the probabilities to find \( P(X \geq 5) \). That is: \[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) \] Evaluate each term and add them together to find the total probability.
07
Compute and conclude
Compute each binomial probability and sum them to get the final result. This will give us the probability that Jessica wins more games than Susan.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Distribution
The binomial distribution is a specific statistical distribution representing the number of successes in a fixed number of independent experiments, where each experiment has a constant probability of success. It is widely used when we are interested in determining the likelihood of a certain number of successes out of a predetermined number of trials. In the context of our exercise, the number of games Jessica wins is modeled using a binomial distribution:
- Each game played is considered as a trial.
- The probability of winning any single game is constant, at 0.4 for Jessica.
- The total number of games played is set at 9, which is the parameter "n" for the distribution.
Probability Calculation
The probability calculation in a binomial distribution is done using the binomial probability formula. This formula allows us to calculate the probability of observing exactly "k" successes out of "n" trials, with the probability of success given as "p". The formula is:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Here, \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose "k" successes from "n" trials. In the exercise, we need to determine the probabilities for Jessica winning 5 or more games out of 9.
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Here, \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose "k" successes from "n" trials. In the exercise, we need to determine the probabilities for Jessica winning 5 or more games out of 9.
- First, calculate the probability for each possible number of games won, from 5 to 9.
- Add these probabilities to get the total probability of winning at least 5 games.
Random Variables
In statistics, a random variable is a quantitative variable whose outcomes are determined by the result of a random phenomenon. In the exercise, the random variable \( X \) is defined as the number of games Jessica wins. Random variables can be classified into two types: discrete and continuous. In our exercise, \( X \) is a discrete random variable, as it can only take on a finite number of values (0 through 9, representing the number of wins).
- A discrete random variable sums probabilities for each possible outcome.
- The binomial distribution is particularly helpful as it provides a structured way to work with such random variables.
- By understanding \( X \) and how it functions in this scenario, one can effectively calculate the likelihood of specific outcomes.
