Problem 39 - Calculus for Biology and Medicine

Question

An urn contains three blue and two white balls. You draw a ball at random, note its color, and replace it. You repeat these steps three times. Let \(X\) denote the total number of white balls. Find \(P(X \leq 1)\).

Step-by-Step Solution

Answer

The probability that at most one white ball is drawn is \( \frac{81}{125} \).

Step 1: Understanding the Problem

We have an urn with 3 blue and 2 white balls. We draw a ball, note its color, and then replace it. We repeat this process three times. We want to find the probability that the number of white balls drawn is at most one, i.e., either 0 or 1 white balls are drawn.

Step 1: Calculate the Probability of Drawing a White Ball

Since there are 3 blue balls and 2 white balls, the total number of balls is 5. The probability of drawing a white ball in a single draw is the ratio of white balls to the total number of balls: \( \frac{2}{5} \).

Step 2: Probability of Drawing Zero White Balls

If no white ball is drawn, then all drawn balls must be blue. The probability of drawing a blue ball is \( \frac{3}{5} \). Therefore, the probability of drawing blue three times in a row (zero white balls) is: \( \left( \frac{3}{5} \right)^3 = \frac{27}{125} \).

Step 3: Probability of Drawing One White Ball

The scenario of drawing exactly one white ball involves choosing one position out of the three trials to be the white ball. The probability for each case is: draw a white ball once \( \frac{2}{5} \) and blue balls twice \( \frac{3}{5} \). Using combinations, the probability for exactly one white ball is: \( 3 \times \left( \frac{2}{5} \right) \times \left( \frac{3}{5} \right)^2 = 3 \times \frac{2}{5} \times \frac{9}{25} = \frac{54}{125} \).

Step 4: Calculate P(X ≤ 1)

To find the probability that the number of white balls drawn is at most one, we add the probability of 0 white balls and 1 white ball: \[ P(X \leq 1) = \frac{27}{125} + \frac{54}{125} = \frac{81}{125} \].

Key Concepts

Combinatorics

In probability, combinatorics helps us understand all possible ways an event can happen. It's about counting combinations or arrangements of items. In the given exercise, we use combinatorics to determine the probability of drawing exactly one white ball in three trials.
When we talk about combinations, we focus on selections where the order doesn't matter. For instance, the problem involves drawing three balls, and we must decide in how many ways we can choose one instance to draw a white ball amidst drawing mostly blue balls.
To simplify, consider each drawing as a position or slot. Placing a white ball in one of these positions involves choosing one slot out of three. This is where combinations come into play, specifically calculated using the formula: \[ C(n, k) = \frac{n!}{k!(n-k)!}\], where \(n\) is the total number of trials, and \(k\) is how many you choose, which in this problem is \(3\) choose \(1\).

• This formula helps to calculate how many possible ways there are to choose the position of the white ball.
• Here, the number of ways to choose one slot from three, gives us \(C(3, 1) = 3\).

So, for every individual way to choose a white ball, we need to combine it with the likelihood of the other two chosen being blue to find the correct probability.

Random Variables

A random variable is a way to quantify outcomes of a random process with numbers. In our exercise, the random variable under consideration is \(X\), which represents the total number of white balls you draw over three attempts.
Imagine every time you draw a ball, you note down whether it is white or blue. Each of these outcomes can be translated into our random variable \(X\). The goal is to understand various possible outcomes of this random process, specifically finding how often \(X\) can take on certain values.

• In our scenario, \(X\) can range from 0 to 3, signifying zero to three white balls being drawn
• Calculating probabilities for these outcomes helps us understand how likely a specific number of white balls appear in the draws.

The ultimate aim is to find \(P(X \leq 1)\), which means calculating the chances that you draw at most one white ball in three draws. Thus, this value of \(X\) provides insight into the distribution of possible outcomes of drawing balls and their probabilities.

Binomial Distribution

Binomial distribution is a probability distribution that describes outcomes of a fixed number of repeated trials. Each trial is independent, and there are only two possible outcomes, typically termed success and failure. In our problem, each draw of a ball represents a trial.
In each trial, drawing a white ball is considered a success, while drawing a blue ball is a failure. Given there are three trials, you repeat the process under identical conditions. The result can be modeled using a binomial distribution where the number of trials \(n\) is 3, and the probability \(p\) of drawing a white ball (success) is \(\frac{2}{5}\).
The binomial distribution formula \[ P(X = k) = C(n, k) \times p^k \times (1-p)^{n-k} \] describes how we calculate the probability of getting a specific number of successes (white balls) in \(n\) trials.

• Here, \(C(n, k)\) is the combination function denoting the number of ways to achieve \(k\) successes in \(n\) trials.
• The expression \(p^k\) represents the likelihood of \(k\) successes, while \((1-p)^{n-k}\) denotes the probability of failures in the remaining trials.

This distribution helps you find \(P(X \leq 1)\), sum of probabilities of drawing zero or one white ball, by finding \(P(X = 0)\) and \(P(X = 1)\), and adding them together to obtain the final probability result.