Problem 84 - A First Course in Probability

Question

An urn contains 12 balls, of which 4 are white. Three players \(-A, B,\) and \(C-\) successively draw from the urn, \(A\) first, then \(B\), then \(C\), then \(A\), and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player if (a) each ball is replaced after it is drawn; (b) the balls that are withdrawn are not replaced.

Step-by-Step Solution

Answer

For the case when balls are replaced after being drawn, the probabilities of winning for players A, B, and C are \(\frac{9}{19}, \frac{6}{19}\), and \(\frac{4}{19}\), respectively. When the balls are not replaced, the probabilities of winning for players A, B, and C are \(\frac{1}{3}, \frac{8}{33}\), and \(\frac{28}{99}\), respectively.

Step 1: Probability of Player A winning in one turn

Since there are 4 white balls and 12 balls in total, the probability of player A drawing a white ball in a single turn is \(\frac{4}{12} = \frac{1}{3}\).

Step 2: Probability of Player B winning in one turn

Player B can only win if player A loses in their turn. The probability of player A losing in one turn is \(1 - \frac{1}{3} = \frac{2}{3}\). If A loses, then the probability of player B drawing a white ball is still \(\frac{1}{3}\). Thus, the probability of player B winning in one turn is \(\frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}\).

Step 3: Probability of Player C winning in one turn

Player C can only win if both player A and B lose in their turns. The probability of player A and B both losing in one turn is \(\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}\). If both A and B lose, the probability of player C drawing a white ball is \(\frac{1}{3}\). Thus, the probability of player C winning in one turn is \(\frac{4}{9} \cdot \frac{1}{3} = \frac{4}{27}\).

Step 4: Finding the probability of each player winning

Now that we have the probabilities for each player winning in one turn, we can find the probability of each player winning by considering the geometric series. Let the probabilities for player A, B, and C winning be \(P_A, P_B\), and \(P_C\), respectively. For player A: \(P_A = \frac{1}{3} + \frac{2}{9}\cdot\frac{4}{27} + \left(\frac{2}{9}\cdot\frac{4}{27}\right)^2 + \ldots\) For player B: \(P_B = \frac{2}{9} + \frac{2}{9}\cdot\frac{4}{27} + \left(\frac{2}{9}\cdot\frac{4}{27}\right)^2 + \ldots\) For player C: \(P_C = \frac{4}{27} + \left(\frac{2}{9}\cdot\frac{4}{27}\right)^2 + \ldots\) The geometric series can be used to calculate the sum, and the sum can be represented as \(\frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio in each series. For player A: \(P_A = \frac{\frac{1}{3}}{1-\frac{8}{27}} = \frac{9}{19}\) For player B: \(P_B = \frac{\frac{2}{9}}{1-\frac{8}{27}} = \frac{6}{19}\) For player C: \(P_C = \frac{\frac{4}{27}}{1-\frac{8}{27}} = \frac{4}{19}\) So, the probabilities of winning for players A, B, and C are \(\frac{9}{19}, \frac{6}{19}\), and \(\frac{4}{19}\) respectively when the balls are replaced after being drawn. (b) Balls are not replaced after being drawn

Step 1: Probability of Player A winning in one turn

With the balls not replaced, the probability of player A drawing a white ball in one turn is the same as before, \(\frac{4}{12} = \frac{1}{3}\).

Step 2: Probability of Player B winning in one turn

Player B can only win if player A loses in their turn. The probability of player A losing in one turn is \(1 - \frac{1}{3} = \frac{2}{3}\). If A loses, there are now 11 balls left, and the probability of player B drawing a white ball is \(\frac{4}{11}\). Thus, the probability of player B winning in one turn is \(\frac{2}{3} \cdot \frac{4}{11} = \frac{8}{33}\).

Step 3: Probability of Player C winning in one turn

Player C can only win if both player A and B lose in their turns. The probability of player A and B both losing in one turn is \(\frac{2}{3} \cdot \frac{7}{11} = \frac{14}{33}\). If both A and B lose, there are now 10 balls left, and the probability of player C drawing a white ball is \(\frac{4}{10} = \frac{2}{5}\). Thus, the probability of player C winning in one turn is \(\frac{14}{33} \cdot \frac{2}{5} = \frac{28}{165}\).

Step 4: Calculate probability of each player winning using conditional probabilities

Let the probabilities for player A, B, and C winning be \(P_A, P_B\), and \(P_C\), respectively. For player A: \(P_A = \frac{1}{3}\) For player B: \(P_B = \frac{8}{33}\) For player C: \(P_C = 1 - P_A - P_B = 1 - \frac{1}{3} - \frac{8}{33} = \frac{28}{99}\) So, the probabilities of winning for players A, B, and C are \(\frac{1}{3}, \frac{8}{33}\), and \(\frac{28}{99}\) respectively when the balls are not replaced after being drawn.

Key Concepts

Geometric series in probability scenarios

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the "common ratio." In probability scenarios like the urn problem, geometric series can help us determine the probability of sequential events.
Suppose player A starts and draws a white ball, the game ends immediately. But if not, we continue with players B and C sequentially. Each player has a defined probability of winning on their turn, and if not, the cycle repeats.

This is where geometric series become handy. Since player A has the first chance, their probability includes terms like \(\frac{1}{3}, \left(\frac{2}{9} \times \frac{4}{27}\right), \left(\frac{2}{9} \times \frac{4}{27}\right)^2, \ldots\) reflecting their chance in the subsequent cycles, forming a geometric series.
To solve a geometric series, we can use the formula \( S = \frac{a}{1-r} \), where \(a\) is the first term, and \(r\) is the common ratio. It systematizes handling probabilities that involve repeating trials, like rounds in the urn exercise.

Conditional probability in sequential draws

Conditional probability gives us the likelihood of an event occurring, given that another event has already happened. It is particularly useful in scenarios like the urn game, where each player's chance of winning is affected by the draws before theirs.
Imagine player A does not draw a white ball initially, changing the conditions for player B. We call the original probability the "prior probability," and the updated probability after considering additional information is the "conditional probability."

• For player B to win, player A must lose, which changes player B's probability to consider this scenario only. Hence, if player A misses, player B's probability is based on the remaining balls in the urn.
• Similarly, player C’s conditional probability of winning depends on both player A and B not winning in their turns.

Understanding conditional probability is essential in such chained events, as it helps calculate each person's true chance of winning after others' actions are factored in. It offers a step-by-step adjustment to the probabilities based on remaining prospects.

Combinatorics and its role in calculating outcomes

Combinatorics is the branch of mathematics concerned with counting, combination, and permutation of sets of elements. In probability problems, it helps determine the number of possible outcomes or arrangements.
In our urn example, the drawing of balls without replacement illustrates combinatorial principles, as the total number of balls decreases with each drawn. This change directly affects the calculation of probabilities, as it modifies combinations possible for future draws.

• When calculating probabilities of drawing white balls, knowing how many ways a white ball can be selected from the remaining is essential.
• Understanding combinations helps in structure scenarios where order doesn’t matter, influencing exact probabilities of events as players draw sequentially from a diminishing set.

By applying combinatorics, we are able to account effectively for outcomes that involve choosing subsets from a set (like picking balls), an unavoidable part of calculating probabilities in draws without replacement.