Problem 7 - Calculus for Biology and Medicine

Question

You draw 3 cards from a standard deck of 52 cards without replacement. Let \(X\) denote the number of spades in your hand. Find the probability mass function describing the distribution of \(X\).

Step-by-Step Solution

Answer

The probability mass function for \(X\) is: \(P(X = 0) = \frac{9,139}{22,100}, P(X = 1) = \frac{9,633}{22,100}, P(X = 2) = \frac{3,042}{22,100}, P(X = 3) = \frac{286}{22,100}\).

Step 1: Define the Random Variable

The random variable \(X\) represents the number of spades in a hand of 3 cards drawn without replacement from a standard deck of 52 cards. Hence, \(X\) can take the values \(0, 1, 2,\) or \(3\).

Step 2: Find Total Number of Outcomes

Calculate the total number of ways to draw 3 cards from a 52-card deck. This is given by the combination \(\binom{52}{3}\). Compute this value: \[\binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,100.\]

Step 3: Calculate Probability of No Spades (\(X = 0\))

If no spades are drawn, all 3 cards must come from the other 39 cards (since there are 13 spades). Calculate the number of ways to choose 3 cards from these 39: \[\binom{39}{3} = \frac{39 \times 38 \times 37}{3 \times 2 \times 1} = 9,139.\] The probability is \(\frac{9,139}{22,100}\).

Step 4: Calculate Probability of One Spade (\(X = 1\))

For one spade, choose 1 spade from 13 and 2 cards from 39: \[\binom{13}{1} \times \binom{39}{2} = 13 \times \binom{39}{2} = 13 \times 741 = 9,633.\] The probability is \(\frac{9,633}{22,100}\).

Step 5: Calculate Probability of Two Spades (\(X = 2\))

For two spades, choose 2 spades from 13 and 1 card from 39: \[\binom{13}{2} \times \binom{39}{1} = 78 \times 39 = 3,042.\] The probability is \(\frac{3,042}{22,100}\).

Step 6: Calculate Probability of Three Spades (\(X = 3\))

All 3 cards are spades. The number of ways to choose all 3 spades from 13 is \[\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286.\] The probability is \(\frac{286}{22,100}\).

Step 7: Compose the Probability Mass Function

The probability mass function for \(X\) is: * \(P(X = 0) = \frac{9,139}{22,100}\)* \(P(X = 1) = \frac{9,633}{22,100}\)* \(P(X = 2) = \frac{3,042}{22,100}\)* \(P(X = 3) = \frac{286}{22,100}\).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and combination of elements within a set. In the context of drawing cards, combinatorics helps us understand the different ways we can select cards from a deck. For example, when drawing 3 cards from a 52-card deck, we use combinations instead of permutations because the order in which cards are drawn doesn't matter. The formula for combinations is given by \( \binom{n}{k} \), which reads as "n choose k". This formula is computed as:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Here, \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. For instance, to find the total number of ways to draw 3 cards from 52, we calculate

• \( \binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,100 \)

This result shows us that there are 22,100 possible combinations of 3 cards from a 52-card deck.

Random Variable

A random variable is a numerical description of the outcome of a random process. In this exercise, the random variable \( X \) represents the number of spades in a hand of 3 cards drawn from a deck without replacement. This means that \( X \) can take on any of the possible outcomes:

• 0 spades,
• 1 spade,
• 2 spades, or
• 3 spades.

These values signify the range over which \( X \) can vary. Understanding random variables helps in setting up probability calculations for different events. Each outcome or value that the random variable can take has a corresponding probability that describes the likelihood of that event occurring. The collection of these probabilities forms the probability mass function.

Card Probability

Card probability refers to the likelihood of various outcomes when drawing cards from a deck. Like in our exercise, where we assess the likelihood of drawing 0, 1, 2, or 3 spades. The key to finding these probabilities lies in considering the number of favorable outcomes divided by the total number of possible outcomes. For example, to find the probability of drawing exactly 2 spades (\( X = 2 \)) from a deck:

• First, we need to calculate the favorable ways: choose 2 spades from 13 and 1 card from the remaining 39. This is calculated as \( \binom{13}{2} \times \binom{39}{1} = 78 \times 39 = 3,042 \).
• Then, divide by the total number of combinations: \( \frac{3,042}{22,100} \).

Perform similar calculations for other possible outcomes. Card probabilities are essential in games of chance, allowing players to make informed decisions based on the likelihood of various hands or draws.