Question

Suppose we make draws from an urn containing two red balls and three black ones, replacing the chosen ball after each draw. How many draws should we make (we have to decide this number in advance) to have probability 0.5 or greater of selecting at least two red balls?

Short answer

5 draws are needed.

Step-by-step solution

Understanding the Probability

The probability of drawing a red ball in one draw is the number of red balls over the total number of balls: \( \frac{2}{5} \). Similarly, the probability of not drawing a red ball (drawing a black ball) is \( \frac{3}{5} \).

Defining the Desired Probability for Multiple Draws

We are interested in finding the number of draws, \( n \), needed to have a probability of at least 0.5 of drawing at least two red balls. This is, we want \( P(\text{at least 2 red balls in } n \text{ draws}) \geq 0.5 \).

Calculating the Probability of Complement Event

The probability of not drawing at least two red balls is the complementary event: drawing zero or one red ball. For zero red balls, the probability is \( \left(\frac{3}{5}\right)^n \). For one red ball, it is \( n \times \left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right)^{n-1} \).

Setting Up the Inequality

The inequality for at least two red balls is: \[1 - \left[\left(\frac{3}{5}\right)^n + n \times \left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right)^{n-1}\right] \geq 0.5\].

Solving the Inequality

Simplify the inequality from Step 4:\[\left(\frac{3}{5}\right)^n + n \times \left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right)^{n-1} \leq 0.5\]Try different values of \( n \) to satisfy this condition.

Testing Values of n

Let’s test \( n = 4 \):- Probability of 0 red balls: \( \left( \frac{3}{5} \right)^4 = \frac{81}{625} \approx 0.1296 \).- Probability of 1 red ball: \( 4 \times \frac{2}{5} \times \left( \frac{3}{5} \right)^3 = \frac{324}{625} \approx 0.5184 \). - Total for 0 or 1 red balls: \( 0.1296 + 0.5184 = 0.648 \).This does not satisfy our condition. Try \( n = 5 \): - Probability for 0 red balls: \( \left( \frac{3}{5} \right)^5 = \frac{243}{3125} \approx 0.0778 \).- Probability for 1 red ball: \( 5 \times \frac{2}{5} \times \left( \frac{3}{5} \right)^4 = \frac{972}{3125} \approx 0.3101 \).- Total: \( 0.0778 + 0.3101 = 0.3879 \).

Verifying Sufficient Draws

With \( n = 5 \), the probability of zero or one red ball is approximately \( 0.3879 \), so the probability of at least two red balls is \( 1 - 0.3879 = 0.6121 \), which is greater than 0.5. Thus, \( n = 5 \) draws are sufficient.

Key concepts

Complementary Events

In probability theory, complementary events are a pair of outcomes where the sum of their probabilities is equal to 1. If an event happens, its complement represents when it does not happen. For example, in the urn problem, the event of drawing at least two red balls in multiple draws has a complement: drawing zero or only one red ball.

The complement of drawing at least two red balls is useful because calculating the probability of zero or one red ball can be simpler. Once this probability is found, subtracting it from 1 gives the probability of the event we are interested in. In mathematical terms, if A is our event of interest, then the probability of A is: \[ P(A) = 1 - P(A^c) \] where \( P(A^c) \) is the probability of the complementary event. This relationship helps solve the given problem by focusing on what is likely not to happen first, and then deriving what we want.

Binomial Distribution

The binomial distribution is crucial in understanding problems involving repeated independent events with two possible outcomes, like drawing a ball with replacement. In our urn scenario, each draw is a trial where we can either draw a red (success) or black (failure) ball.

To model the probability of drawing at least two red balls in multiple draws, we use the binomial distribution. If \( n \) is the number of draws, \( p \) the probability of drawing a red ball in one draw, the distribution describes the probability of achieving a certain number of successes (red balls) in \( n \) trials. For our problem, \( p = \frac{2}{5} \) per draw.

The binomial probability formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes in \( n \) trials. Understanding this helps us determine the probability for different numbers of red balls drawn.

Inequality

Inequalities in probability help us determine when our solutions reach desired levels. Here, the goal was to have a probability of at least 0.5 for drawing at least two red balls. Formed as: \[ P(\text{at least 2 red balls}) \geq 0.5 \] an effective approach uses complement principles. We set an inequality: \[ \left(\frac{3}{5}\right)^n + n \times \left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right)^{n-1} \leq 0.5 \] representing less than two successes (red balls) over \( n \) trials.

Solving this inequality involves testing different integer values of \( n \) to find the smallest \( n \) satisfying the condition. For the example, we find that 5 draws meet our target, as this results in a probability exceeding 0.5 for drawing at least two red balls. Inequalities can thus help us determine effective strategies in probabilistic scenarios.