Question

There are two boxes. Box I contains 4 Red and 3 white balls. Box II contains 5 red and 2 white balls. Two balls are transferred from Box I to Box II. One ball is then drawn from box II randomly. What is the probability for that ball to be red? (a) \((43 / 63)\) (b) \((23 / 73)\) (c) \((34 / 63)\) (d) None

Short answer

The probability for the ball to be red is \(\frac{43}{63}\).

Step-by-step solution

Define the Events

Let's denote the various events as follows: Event A1: Both transferred balls are red Event A2: One transferred ball is red and the other is white Event A3: Both transferred balls are white Event B: Drawing a red ball from Box II We need to find the probability P(B).

Calculate Probabilities of Events A1, A2, and A3

- To find P(A1), we choose 2 red balls from the 4 available in Box I, and divide by the total number of ways to choose 2 balls out of 7: \(P(A1) = \frac{\binom{4}{2}}{\binom{7}{2}} = \frac{6}{21}\) - To find P(A2), we choose 1 red ball from the 4 available and 1 white ball from the 3 available, then divide by the total number of ways to choose 2 balls: \(P(A2) = \frac{\binom{4}{1} \binom{3}{1}}{\binom{7}{2}} = \frac{12}{21}\) - To find P(A3), we choose 2 white balls from the 3 available in Box I, and divide by the total number of ways to choose 2 balls out of 7: \(P(A3) = \frac{\binom{3}{2}}{\binom{7}{2}} = \frac{3}{21}\)

Calculate P(B|A1), P(B|A2), and P(B|A3)

- P(B|A1): Both transferred balls are red, so now Box II has 7 red balls and 2 white balls. The probability of drawing a red ball is: \(P(B|A1) = \frac{7}{9}\) - P(B|A2): One red and one white ball are transferred, so Box II has 6 red balls and 3 white balls. The probability of drawing a red ball is: \(P(B|A2) = \frac{6}{9}\) - P(B|A3): Both transferred balls are white, so Box II has 5 red balls and 4 white balls. The probability of drawing a red ball is: \(P(B|A3) = \frac{5}{9}\)

Use Conditional Probability Formula to Calculate P(B)

Now, we can use conditional probability to calculate the probability of drawing a red ball from Box II as follows: \(P(B) = P(B|A1)P(A1) + P(B|A2)P(A2) + P(B|A3)P(A3)\) Plugging in the probabilities we calculated in Steps 2 and 3, we get: \(P(B) = (\frac{7}{9})(\frac{6}{21}) + (\frac{6}{9})(\frac{12}{21}) + (\frac{5}{9})(\frac{3}{21})\) \(P(B) = \frac{42 + 72 + 15}{63}\) \(P(B) = \frac{129}{63}\) Since this probability is not in its simplest form, we will simplify it: \(P(B) = \frac{43}{63}\) Thus, the answer is (a).

Key concepts

Conditional Probability

Conditional probability is a key concept when dealing with events that are affected by previous outcomes. It helps us determine the probability of an event occurring given that another event has already occurred. This concept is essential in solving problems involving multiple stages, like the transfer of balls between boxes before drawing one.In the exercise, we calculated the probability of drawing a red ball from Box II after transferring two balls from Box I. The probability of drawing a red ball depends on the colors of the balls transferred, which is determined by events A1, A2, and A3.To calculate the overall probability of drawing a red ball from Box II, we use the formula for conditional probability: \[ P(B) = P(B|A1)P(A1) + P(B|A2)P(A2) + P(B|A3)P(A3) \] Each term in the formula represents the probability of drawing a red ball under a specific scenario of transferred balls, multiplied by the probability of that scenario occurring. This method ensures that all potential outcomes are considered and weighted appropriately.

Combinatorics

Combinatorics is the mathematics of counting. It helps us determine how many ways we can arrange or select items from a set, which is essential when calculating probabilities. This tool was fundamental in solving the exercise since it allowed us to calculate the different probabilities of transferring balls between the boxes.In our problem, we used combinatorics to calculate the number of ways two balls can be chosen from Box I in different color combinations. The formula for combinations is:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

• Choosing 2 red balls (Event A1): Utilized \( \binom{4}{2} \).
• Choosing 1 red and 1 white ball (Event A2): Utilized \( \binom{4}{1} \binom{3}{1} \).
• Choosing 2 white balls (Event A3): Utilized \( \binom{3}{2} \).

These calculations let us determine the probabilities of each event occurring given the total possibilities.

Event Calculation

Event calculation involves the determination and analysis of events and their probabilities. In this specific situation, the events were related to the way balls were transferred between the boxes and the implication of each scenario on the final outcome of drawing a red ball. First, we defined distinct events based on possible transfers using labels A1, A2, and A3. This step helped in organizing our approach to finding probabilities efficiently.

• Event A1: Presentation of both balls being red leads to a new configuration in the second box.
• Event A2: Presentation of one red and one white ball alters proportions differently in the second box.
• Event A3: Presentation of both white balls modifies yet another distinct setup.

Once these events were clearly established and the probabilities were calculated, we used them to respect the conditional nature of our final probability query. Each event affects the composition of Box II and thus the likelihood of different outcomes when a ball is drawn. This systematic event calculation aids immensely in logically deducing the final probability results efficiently and accurately.