Problem 24 - A First Course in Probability

Question

3.24. Each of 2 balls is painted either black or gold and then placed in an urn. Suppose that each ball is colored black with probability \(\frac{1}{2}\) and that these events are independent. (a) Suppose that you obtain information that the gold paint has been used (and thus at least one of the balls is painted gold). Compute the conditional probability that both balls are painted gold. (b) Suppose now that the urn tips over and 1 ball falls out. It is painted gold. What is the probability that both balls are gold in this case? Explain.

Step-by-Step Solution

Answer

The short answer for this problem is: (a) The conditional probability that both balls are painted gold, given that at least one ball is painted gold, is \(\frac{1}{3}\). (b) The conditional probability that both balls are painted gold, given that the first ball observed is gold, is \(\frac{1}{2}\).

Step 1: Calculate P(A)

To start, we need to find the probability of both balls being painted gold (event A). Since each ball has a \(\frac{1}{2}\) probability of being painted black or gold, and the events are independent, to find P(A), we simply square the probability of painting one ball gold: \[P(A) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\]

Step 2: Calculate P(B1)

We want the probability that at least one ball is painted gold. We can find this by calculating the probability that both balls are painted black and then subtract that probability from 1, as they are complementary events: \[P(B1) = 1 - P(\text{both balls black})\] \[P(\text{both balls black}) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\] \[P(B1) = 1 - \frac{1}{4} = \frac{3}{4}\]

Step 3: Calculate P(A ∩ B1)

Since A is a subset of B1 (if both balls are gold, it's certain that at least one ball is gold), we have: \[P(A \cap B1) = P(A) = \frac{1}{4}\]

Step 4: Calculate the conditional probability P(A|B1)

Now, we plug the probabilities into the formula for conditional probability: \[P(A|B1) = \frac{P(A \cap B1)}{P(B1)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}\] Thus, the conditional probability that both balls are painted gold, given that at least one ball is painted gold, is \(\frac{1}{3}\). **(b) Calculate P(A|B2)** Now, we need to compute the conditional probability P(A|B2). We will use the same formula: \[P(A|B2) = \frac{P(A \cap B2)}{P(B2)}\]

Step 1: Calculate P(B2)

First, we need to find the probability that the first ball observed is gold. Since there's a \(\frac{1}{2}\) chance that either ball is gold, we have: \[P(B2) = \frac{1}{2}\]

Step 2: Calculate P(A ∩ B2)

Now we need to find the probability of both events A and B2 occurring. If both balls are gold, observing one ball as gold is certain: \[P(A \cap B2) = P(A) = \frac{1}{4}\]

Step 3: Calculate the conditional probability P(A|B2)

Now, we plug the probabilities into the formula for conditional probability: \[P(A|B2) = \frac{P(A \cap B2)}{P(B2)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}\] Thus, the conditional probability that both balls are painted gold, given that the first ball observed is gold, is \(\frac{1}{2}\).

Key Concepts

Independent Events

Independent events are a fundamental concept in probability theory. If you are dealing with independent events, the outcome of one event does not affect the outcome of another. When it comes to the exercise, each ball being either black or gold is independent of the other, meaning the color of one ball doesn't influence the color of the other.

In mathematical terms, two events A and B are said to be independent if the probability of both occurring is the product of their individual probabilities:

• If event A occurs with probability \( P(A) \) and event B with probability \( P(B) \), then the probability of both A and B occurring (\( P(A \cap B) \)) is \( P(A) \times P(B) \).
• For example, if both balls being gold is event A and the probability of one ball being gold is \( \frac{1}{2} \), then \( P(A) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

Understanding independent events simplifies complex probability problems as it allows for straightforward calculation of joint probabilities.

Probability Calculation

Probability is the measure of how likely an event is to occur. To calculate probabilities, especially in the context of independent events, we use simple mathematical operations. In our exercise, we need to determine the probability of different combinations of two balls either being black or gold.

Here is a quick guide on calculating these probabilities:

• Start by identifying the probability of each event separately. The probability that one ball is gold is \( \frac{1}{2} \).
• For independent events, calculate the combined probability by multiplying the probabilities of the individual events. For example, for both balls to be gold, the probability is \( \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \).
• For more complex scenarios like determining the probability that at least one ball is gold, consider complementary events, which we will discuss in the next section.

Being able to correctly calculate these probabilities allows you to assess different outcomes and better understand the likelihood of each scenario.

Complementary Events

Complementary events are events that cover all possible outcomes of a random experiment. If one event occurs, the other cannot, and vice versa. When considering the likelihood of at least one ball being gold, thinking about complementary events helps simplify the calculation.

• The probability of both balls being black can be found by squaring the probability of one ball being black: \( \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \).
• The complementary event to both balls being black is having at least one ball gold, which can be found using: \( P(\text{at least one gold}) = 1 - P(\text{both black}) = 1 - \frac{1}{4} = \frac{3}{4} \).

To use complementary events effectively, remember that the probability of an event and its complement always add up to one. This method can often hasten your calculation and serve as a check for your final answer.