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Chapter 8: Problem 27

Calculate the expected value of the given random variable \(X .\) [Exercises \(23,24,27\), and 28 assume familiarity with counting arguments and probability (Section 7.4).] Select five cards without replacement from a standard deck of 52, and let \(X\) be the number of queens you draw.

Short Answer

Expert verified
The expected value of the random variable \(X\) (the number of queens drawn from a standard deck of 52 cards when selecting 5 cards without replacement) is approximately 0.385.

Step by step solution

01

Find the total number of ways to draw 5 cards without replacement

We have a deck of 52 cards and we want to draw 5 cards without replacement. To find the total number of ways to draw 5 cards, we use the combination formula, which is: \[ C(n, k) = \frac{n!}{(n - k)!k!} \] In our case, \(n = 52\) and \(k = 5\). Therefore, the total number of ways is: \[ C(52, 5) = \frac{52!}{(52 - 5)!5!} = \frac{52!}{47!5!} = 2,598,960 \]
02

Find the probabilities of drawing 0, 1, 2, 3, or 4 queens

There are 4 queens and 48 non-queens in a standard deck of 52 cards. To find the probabilities of drawing 0, 1, 2, 3, or 4 queens, we will calculate the combinations for each case and divide that by the total number of combinations calculated in Step 1. Probability of drawing 0 queens: We want to select 5 cards from the non-queens. That gives us: \[ P(X=0) = \frac{C(48, 5)}{C(52, 5)} = \frac{1,712,304}{2,598,960} = 0.658842 \] Probability of drawing 1 queen: We want to select 1 queen and 4 non-queens. That gives us: \[ P(X=1) = \frac{C(4, 1) \times C(48, 4)}{C(52, 5)} = \frac{778,320}{2,598,960} = 0.299278 \] Probability of drawing 2 queens: We want to select 2 queens and 3 non-queens. That gives us: \[ P(X=2) = \frac{C(4, 2) \times C(48, 3)}{C(52, 5)} = \frac{108,024}{2,598,960} = 0.041569 \] Probability of drawing 3 queens: We want to select 3 queens and 2 non-queens. That gives us: \[ P(X=3) = \frac{C(4, 3) \times C(48, 2)}{C(52, 5)} = \frac{2,304}{2,598,960} = 0.000887 \] Probability of drawing 4 queens: It is not possible to select all 4 queens in only 5 cards, so the probability is 0.
03

Calculate the expected value of X

Now we will calculate the expected value of X by multiplying the probabilities calculated in Step 2 by their respective values (0, 1, 2, or 3) and summing them up: \[ E(X) = (0 \times 0.658842) + (1 \times 0.299278) + (2 \times 0.041569) + (3 \times 0.000887) = 0.299278 + 0.083138 + 0.002661 = 0.385077 \] The expected value of the random variable \(X\) (the number of queens drawn from a standard deck of 52 cards when selecting 5 cards without replacement) is approximately 0.385.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combination Formula
The combination formula is a powerful tool in probability theory, particularly in problems involving selections from a group. It is used to determine the number of ways to choose a subset of items from a larger set, where the order of selection does not matter. This is especially relevant when drawing cards from a deck. The combination formula is given by: \[ C(n, k) = \frac{n!}{(n - k)!k!} \] Where:
  • \( n \) is the total number of items in the group.
  • \( k \) is the number of items to choose.
  • \( ! \) denotes a factorial, which is the product of all positive integers up to that number.
In the context of card probability, such as developing strategies in poker or calculating the likelihood of specific hands, the combination formula allows us to calculate the total possible outcomes for different scenarios.
Expected Value
Expected value is a central concept in probability theory, representing the mean value or long-term average of a random variable's outcomes. When calculating the expected value of a random variable, we take into account all possible outcomes and their probabilities. The formula for expected value \( E(X) \) of a discrete random variable \( X \), which can take on values \( x_1, x_2, \ldots, x_n \) each with probability \( P(x_1), P(x_2), \ldots, P(x_n) \) respectively, is: \[ E(X) = \sum_{i=1}^{n} x_i \cdot P(x_i) \] This helps in predicting the expected number of queens in hands of cards drawn. Even with randomness involved, it gives a statistical average that is useful in many practical scenarios, such as betting probabilities or risk analysis.
Random Variables
A random variable is a numerical outcome of a random phenomenon. In this context, the random variable \( X \) represents the number of queens drawn from a hand of cards. Random variables can be either discrete or continuous, but in our specific card example, \( X \) is discrete because it can take on a finite set of values \( \{0, 1, 2, 3\} \). Understanding random variables includes knowing their probability distributions, which describe how probabilities are distributed over the values of the random variable. Various statistical properties, like expected value and variance, can be derived from these distributions. Random variables are pivotal in defining probabilities for specific scenarios.
Card Probability
Card probability deals with the likelihood of drawing specific cards or hands from a deck, particularly important in games like poker or bridge. In a standard 52-card deck, each card draw is influenced by those that have already been drawn. This makes understanding probability slightly complex but fascinating. When drawing without replacement, each draw affects the probabilities of future draws. For example, drawing one queen affects the likelihood of drawing another queen because there are fewer queens remaining in the deck and fewer total cards. When calculating probabilities for hands, it's crucial to consider both the total combination of cards possible and specific characteristics of desired hands. This involves using the combination formula to break down complex probability problems into manageable parts.

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