Problem 73 - Calculus for Biology and Medicine

Question

An urn contains one black and nine white balls. Balls are drawn at random until the black ball is selected. Find the probability that exactly six white balls will be drawn before the black one is if (a) each ball is replaced before the next ball is drawn and (b) balls are not replaced.

Step-by-Step Solution

Answer

With replacement: \( \frac{531441}{10000000} \); without replacement: 0.1.

Step 1: Understanding the Problem

We need to find the probability of drawing exactly six white balls before the black ball. There are two scenarios to consider: when each ball is drawn with replacement and when the ball is drawn without replacement.

Step 2: Calculate Probability with Replacement

In the replacement scenario, each draw is an independent event. The probability of drawing a white ball is \( \frac{9}{10} \) and the probability of drawing a black ball is \( \frac{1}{10} \). We aim for a sequence: six white balls followed by one black ball.

Step 3: Calculate Probability for Replacement Scenario

The probability is calculated by the product of drawing six white balls and one black ball:\[ \text{Probability} = \left( \frac{9}{10} \right)^6 \times \left( \frac{1}{10} \right) = \frac{9^6}{10^7} = \frac{531441}{10000000}. \]

Step 4: Calculate Probability without Replacement

Without replacement, drawing probabilities change. We need exactly six white balls (out of nine) before the black ball is drawn. The sequence changes dynamically as balls are not replaced.

Step 5: Choose Specific Sequence for Non-replacement

Compute the probability of drawing six white balls first from the nine:\[ \text{Probability for first 6 white } = \frac{9}{10} \times \frac{8}{9} \times \frac{7}{8} \times \frac{6}{7} \times \frac{5}{6} \times \frac{4}{5} = 0.4. \]

Step 6: Draw the Black Ball after the Sequence

After these draws, the probability of the next ball being black is:\[ \frac{1}{4}. \]

Step 7: Calculate Overall Probability Without Replacement

Combine steps to get the total probability:\[ \text{Total Probability} = 0.4 \times \frac{1}{4} = 0.1. \]

Key Concepts

Replacement Scenarios

In probability theory, a replacement scenario refers to a situation where each item or event is returned to its original condition after being used. This concept is crucial when dealing with probability experiments. Let's break it down with our urn problem. When you draw a ball, every time it is replaced back before the next draw, maintaining the original ratio of black to white balls.
This implies that each draw is not affected by previous draws. You maintain a constant probability of drawing each type of ball. For example, drawing a black ball has a probability of \( \frac{1}{10} \), and drawing a white ball has a probability of \( \frac{9}{10} \).

• In replacement scenarios, probabilities remain constant.
• No matter how many times a draw occurs, the probabilities do not change.

This makes calculations straightforward as you consistently apply the same probability for each independent event, as seen in our solution with replacement.

Independent Events

Independent events are events where the outcome of one event does not affect the outcome of another. When dealing with probability scenarios involving replacements, every draw from an urn represents an independent event.
In such scenarios, the probability of each event remains unaffected by other events. This is exemplified by the act of drawing six white balls and then a black ball when each ball is replaced.

• The probability of each draw is exactly the same.
• Each event stands alone — the outcome of one does not influence the others.

For calculating probabilities, the independence of events implies multiplying probabilities for a sequence. In this case, draw six times \( \frac{9}{10} \) for white, and once \( \frac{1}{10} \) for black, resulting in the product \( \left( \frac{9}{10} \right)^6 \times \frac{1}{10} \).

Sequential Probability

Sequential probability refers to the calculation of probabilities for a series of events occurring in a specific order. When analyzing the urn problem, each draw contributes to forming a sequence.
For the replacement scenario, we aim for six consecutive white balls followed by a black ball. Each part of this sequence has a fixed probability. To find the total probability of this sequence, multiply the probabilities of each individual event.

• The sequence is planned: six white balls, then a black ball.
• Each draw is treated separately, allowing for straightforward multiplication of probabilities.

Without replacement, however, the sequence of draws requires dynamic recalculation as the situation changes with each draw. Nonetheless, sequencing is crucial as it organizes events to solve the problem.

Dynamic Probability Change

In scenarios without replacement, probabilities change dynamically with each draw. Every time a ball is drawn and not replaced, the total number of balls decreases, altering the probability landscape.
Originally, the probability of drawing the first white ball is \( \frac{9}{10} \), but with each successive draw, as more balls are removed, the probability changes.

• Probability adjustments are necessary after each draw.
• The initial setup shifts, requiring recalculations at each step.

This dynamic change is evident in the urn problem without replacement. After successfully drawing six white balls, the probability of the subsequent event being a black ball changes to \( \frac{1}{4} \). The total sequence probability becomes a product of adjusted steps, calculated as \( 0.4 \times \frac{1}{4} = 0.1 \). Understanding these dynamics is key to solving non-replacement probability problems effectively.