Problem 40 - Calculus for Biology and Medicine

Question

An urn contains four red, seven green, and two white balls. You draw a ball at random, note its color, and replace it. You repeat these steps four times. Let \(X\) denote the number of red balls and \(Y\) the number of green balls. Find \(P(X+Y=2)\).

Step-by-Step Solution

Answer

The probability that \(X+Y=2\) is \(\frac{2904}{28561}\).

Step 1: Understand the problem

We need to find the probability that the combined number of red and green balls drawn equals 2, after drawing 4 balls with replacement. This is denoted as \(X+Y=2\).

Step 2: Calculate individual probabilities

First, determine the probability of drawing a red ball: \( P(R) = \frac{4}{13} \), a green ball: \( P(G) = \frac{7}{13} \), and a white ball: \( P(W) = \frac{2}{13} \).

Step 3: Express \(X+Y\) events

Since \(X+Y = 2\), consider events where two of the draws are either red or green, and the rest are white. Possible combinations are: 2 red and 0 green, 1 red and 1 green, 0 red and 2 green.

Step 4: Calculate probabilities for specific scenarios

Calculate for each case: Case 1: 2 red and 0 green \((2W)\): \(P(X = 2, Y = 0) = \binom{4}{2} \left(\frac{4}{13}\right)^2 \left(\frac{2}{13}\right)^2 = 6 \times \frac{16}{169} \times \frac{4}{169} = \frac{384}{28561}\).Case 2: 1 red and 1 green \((2W)\): \(P(X = 1, Y = 1) = \binom{4}{1} \binom{3}{1} \left(\frac{4}{13}\right) \left(\frac{7}{13}\right) \left(\frac{2}{13}\right)^2 = 12 \times \frac{28}{169} \times \frac{4}{169} = \frac{1344}{28561}\).Case 3: 0 red and 2 green \((2W)\): \(P(X = 0, Y = 2) = \binom{4}{2} \left(\frac{7}{13}\right)^2 \left(\frac{2}{13}\right)^2 = 6 \times \frac{49}{169} \times \frac{4}{169} = \frac{1176}{28561}\).

Step 5: Sum up probabilities

Add the probabilities of each case to get the total probability of \(X+Y=2\): \[P(X+Y=2) = \frac{384}{28561} + \frac{1344}{28561} + \frac{1176}{28561} = \frac{2904}{28561}\].

Key Concepts

Combinatorics

Combinatorics looks at different ways to count combinations and arrangements of items. This technique is useful when calculating probabilities in scenarios like drawing balls from an urn. In the exercise, combinatorics is used to determine how many ways certain events can happen.

You used binomial coefficients to figure out different ways you could draw 2 red balls, 1 red and 1 green ball, or 0 red and 2 green balls out of 4 total draws. The symbol \( \binom{n}{k} \) represents the number of combinations of choosing \(k\) items from \(n\) items without considering the order.

• Case 1: \( \binom{4}{2} \) was used when choosing 2 red balls out of 4 draws.
• Case 2: \( \binom{4}{1} \binom{3}{1} \) helped calculate choosing 1 red and 1 green ball from 4 draws.
• Case 3: \( \binom{4}{2} \) was used for selecting 2 green balls.

These combinations were essential in determining the likelihood of different outcomes that satisfied the condition \(X+Y = 2\). By compiling all possibilities, you gain clarity on how various setups can yield the same result.

Random variables

Random variables simplify your understanding of probability problems by turning complex scenarios into manageable mathematical terms. This exercise uses two random variables: \(X\) and \(Y\).

Random variable \(X\) denotes the number of red balls drawn, while \(Y\) stands for the number of green balls. Rather than viewing each draw separately, random variables let you focus on specific outcomes, reducing complexity.

These variables help in quantifying uncertain outcomes, essential for working with probabilities. Here you focused on cases where the total number of red and green balls (\(X+Y\)) equals 2. Understanding that \(X\) and \(Y\) have discrete values—dependent on the actual draws—forms the core of the analysis. You use probabilities and randomness to anticipate different combinations and prepare for potential results.

Probability distribution

Probability distribution describes how probabilities are spread over possible outcomes. Understanding this can help you determine which scenarios are more likely. Here, you look for combinations of red and green balls drawn in terms of \(X+Y=2\).

Each specific event, such as drawing exactly two red balls or one red and one green ball, corresponds to a probability, forming a probability distribution across all desired possibilities.

• Event 1: 2 red, 0 green gives a probability of \(\frac{384}{28561}\).
• Event 2: 1 red, 1 green results in \(\frac{1344}{28561}\).
• Event 3: 0 red, 2 green corresponds to \(\frac{1176}{28561}\).

The sum of these probabilities provides the total likelihood of \(X+Y\) equating to 2, creating a full picture of what outcomes are probable when random sampling occurs. Studying individual events gives insight into which mixtures of red and green support your condition best.