Problem 40 - A First Course in Probability

Question

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls.

Step-by-Step Solution

Answer

The probabilities of drawing exactly the following number of white balls in a sample of size 3 are: (a) 0 white balls: \(0.3365\) (b) 1 white ball: \(0.4416\) (c) 3 white balls: \(0.1669\) (d) 2 white balls: \(0.0550\)

Step 1: Find the probability of drawing 0 white balls

To have 0 white balls in the sample, we need to draw 3 red balls. The initial probability of drawing a red ball is 7/12. After drawing a red ball, the probability of drawing a red ball again on the second draw becomes 8/13 (since we added a red ball). On the third draw, the probability increases to 9/14. Therefore, the probability of drawing 0 white balls can be calculated as:\( \frac{7}{12} * \frac{8}{13} * \frac{9}{14} \)

Step 2: Calculate the probability

Multiply the fractions, we have the probability of 0 white balls in the sample: \( \frac{7 * 8 * 9}{12 * 13 * 14} \approx 0.3365 \) (b) 1 white ball

Step 3: Find the probability of drawing 1 white ball

To have 1 white ball in the sample, we can have the following sequences: white-red-red (WRR), red-white-red (RWR), or red-red-white (RRW). We need to calculate the probabilities of each sequence and sum them up. For WRR: \( \frac{5}{12} * \frac{7}{13} * \frac{8}{14} \) For RWR: \( \frac{7}{12} * \frac{5}{13} * \frac{8}{14} \) For RRW: \( \frac{7}{12} * \frac{8}{13} * \frac{5}{14} \)

Step 4: Calculate the probability

Summing up the probabilities of each sequence gives the probability of having 1 white ball in the sample: \( \frac{5*7*8 + 7*5*8 + 7*8*5}{12*13*14} \approx 0.4416 \) (c) 3 white balls

Step 5: Find the probability of drawing 3 white balls

To have 3 white balls in the sample, we need to draw 3 white balls. The initial probability of drawing a white ball is 5/12. After the first draw, the probability becomes 6/13. Then, the probability of drawing a white ball on the third draw is 7/14. The probability of drawing 3 white balls can be calculated as:\( \frac{5}{12} * \frac{6}{13} * \frac{7}{14} \)

Step 6: Calculate the probability

Multiply the fractions, we have the probability of 3 white balls in the sample: \( \frac{5 * 6 * 7}{12 * 13 * 14} \approx 0.1669 \) (d) 2 white balls

Step 7: Find the probability of drawing 2 white balls

To have 2 white balls in the sample, we can have the following sequences: white-white-red (WWR), white-red-white (WRW), or red-white-white (RWW). We need to calculate the probabilities of each sequence and sum them up. For WWR: \( \frac{5}{12} * \frac{6}{13} * \frac{7}{14} \) For WRW: \( \frac{5}{12} * \frac{7}{13} * \frac{6}{14} \) For RWW: \( \frac{7}{12} * \frac{5}{13} * \frac{6}{14} \)

Step 8: Calculate the probability

Summing up the probabilities of each sequence gives the probability of having 2 white balls in the sample: \( \frac{5*6*7 + 5*7*6 + 7*5*6}{12*13*14} \approx 0.0550 \) The probabilities for each scenario are: (a) 0 white balls: 0.3365 (b) 1 white ball: 0.4416 (c) 3 white balls: 0.1669 (d) 2 white balls: 0.0550

Key Concepts

Sampling with Replacement

Sampling with replacement is a technique used in probability theory where objects are drawn from a set, and then returned to the set after each draw, allowing them to be chosen more than once. This concept is important in situations where we want to maintain the same sampling conditions for each draw.

• In the exercise, each time we draw a ball, it is returned to the urn with an additional ball of the same color.
• This means the composition of the urn changes slightly with each draw, but the same color balls remain available for future draws.

Drawing with replacement contrasts with drawing without replacement, where objects are not returned, altering the total number of objects and the probabilities of future draws. In our exercise, the probabilities for each subsequent draw depend on the previous outcomes since our sample composition increases with each draw. The concept of sampling with replacement is vital for understanding how probabilities adjust as repeated trials occur. This dynamic setup makes calculating probabilities more complex, as shown by the changing denominators and numerators in the solution.

Conditional Probability

Conditional probability refers to the likelihood of an event occurring, given that another event has already taken place. It helps us understand how prior events can influence the probability of subsequent events.

• For example, in our exercise, the probability of drawing a white ball on the second draw depends on whether the first ball drawn was white or red.
• If the first ball is white, another white ball is added to the urn, altering the proportion of white to red balls and affecting the subsequent probabilities.

Understanding conditional probability is essential when outcomes are interdependent. In the exercise, each draw affects the conditions under which the next draw is made, illustrating the principle of 'conditioning' the probability on past events. In mathematical terms, if the probability of event A is influenced by event B, it can be denoted as \( P(A|B) \), which reads as "the probability of A given B." This is calculated by taking the joint probability of A and B, \( P(A \cap B) \), and dividing it by the probability of B, \( P(B) \). This nuanced calculation is vital in problems such as this exercise where sequential events influence each other.

Combinatorics

Combinatorics is the field of mathematics that deals with counting and arranging possibilities in a structured manner, and it is pivotal in calculating probabilities. This field helps identify all potential sequences of events, as seen in our problem.

• Consider the different ways to draw one white ball among three draws: WRR, RWR, and RRW.
• Each sequence has its distinct probability calculation based on the sample path (the sequence of colors drawn).

Combinatorics leverages principles like the multiplication rule to determine the number of possible paths or outcomes. The notion "order matters" means that each sequence must be evaluated separately if the order of events affects the outcome, as different orders can lead to different probabilities. For example, WRR has a different probability than RWR due to different stages of draws impacting the urn's content. Using combinatorial methods allows us to systematically compute probabilities for complex, multi-stage events by considering all possible outcomes. It provides a structured way to handle large and intricate probability spaces, reducing the chance of miscalculations or oversight.