Understanding the Probability Problem

This question asks for the probability of drawing a white ball when we first randomly choose one of two bags and then draw a ball from that chosen bag. This involves concepts of conditional probability and the law of total probability.

Analyzing the Contents of Each Bag

We are given information about the contents of two distinct bags:

• Bag 1: Contains 3 white balls and 2 black balls.
• Bag 2: Contains 5 white balls and 3 black balls.

Let's calculate the total number of balls in each bag:

• Total balls in Bag 1 = 3 (white) + 2 (black) = 5 balls.
• Total balls in Bag 2 = 5 (white) + 3 (black) = 8 balls.

Calculating Probabilities

First, a bag is chosen at random. Since there are two bags, the probability of choosing either bag is equal.

• Probability of choosing Bag 1, denoted as \(P(\text{Bag 1})\) is \(\frac{1}{2}\).
• Probability of choosing Bag 2, denoted as \(P(\text{Bag 2})\) is \(\frac{1}{2}\).

Next, we need to consider the probability of drawing a white ball given which bag was chosen. This is conditional probability.

• Probability of drawing a white ball given Bag 1 was chosen, denoted as \(P(\text{White} | \text{Bag 1})\) is the number of white balls in Bag 1 divided by the total balls in Bag 1:

  \( P(\text{White} | \text{Bag 1}) = \frac{\text{Number of white balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{3}{5} \)

• Probability of drawing a white ball given Bag 2 was chosen, denoted as \(P(\text{White} | \text{Bag 2})\) is the number of white balls in Bag 2 divided by the total balls in Bag 2:

  \( P(\text{White} | \text{Bag 2}) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{5}{8} \)

To find the overall probability of drawing a white ball, \(P(\text{White})\), we use the Law of Total Probability. This law states that the total probability of an event is the sum of its probabilities under each possible condition, weighted by the probability of each condition.

In this case, the event is drawing a white ball, and the conditions are which bag is chosen. The formula is:

\( P(\text{White}) = P(\text{White} | \text{Bag 1}) \times P(\text{Bag 1}) + P(\text{White} | \text{Bag 2}) \times P(\text{Bag 2}) \)

Let's substitute the values we calculated:

\( P(\text{White}) = \left(\frac{3}{5}\right) \times \left(\frac{1}{2}\right) + \left(\frac{5}{8}\right) \times \left(\frac{1}{2}\right) \)

Performing the multiplication:

\( P(\text{White}) = \frac{3 \times 1}{5 \times 2} + \frac{5 \times 1}{8 \times 2} = \frac{3}{10} + \frac{5}{16} \)

To add these fractions, we need a common denominator. The least common multiple (LCM) of 10 and 16 is 80.

• Convert \(\frac{3}{10}\) to a fraction with denominator 80:

  \( \frac{3}{10} = \frac{3 \times 8}{10 \times 8} = \frac{24}{80} \)

• Convert \(\frac{5}{16}\) to a fraction with denominator 80:

  \( \frac{5}{16} = \frac{5 \times 5}{16 \times 5} = \frac{25}{80} \)

Now, add the converted fractions:

\( P(\text{White}) = \frac{24}{80} + \frac{25}{80} = \frac{24 + 25}{80} = \frac{49}{80} \)

Final Probability Result

The probability of drawing a white ball after choosing a bag at random is \(\frac{49}{80}\).

Probability Revision Table

Additional Probability Information

This problem is a classic example demonstrating how to apply probability rules when dealing with multiple stages or conditions. Here are some key concepts related to this problem:

• Conditional Probability: The probability of an event occurring given that another event has already occurred. It's denoted as $P(A|B)$, which is the probability of event A happening given that event B has happened. In our problem, \(P(\text{White} | \text{Bag 1})\) is a conditional probability.
• Law of Total Probability: Used to find the marginal probability of an event by considering all the possible mutually exclusive conditions under which the event can occur. If \(B_1, B_2, ..., B_n\) are mutually exclusive and exhaustive events (meaning one of them must happen), then for any event A:

  \( P(A) = \sum_{i=1}^{n} P(A|B_i) P(B_i) \)

  In our case, the mutually exclusive and exhaustive events are choosing Bag 1 or choosing Bag 2.
• Random Selection: When something is chosen at random from a set of options, each option has an equal probability of being selected. Here, choosing a bag at random means each bag has a probability of \(\frac{1}{\text{Number of bags}}\) of being chosen.

Understanding these concepts is crucial for solving multi-stage probability problems like drawing balls from bags or other scenarios involving sequential or conditional events.