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Chapter 18: Problem 766

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

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

Step by step solution

01

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).
02

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: \(P(A3) = \frac{\binom{3}{2}}{\binom{7}{2}} = \frac{3}{21}\)
03

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: \(P(B|A3) = \frac{5}{9}\)
04

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).

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Most popular questions from this chapter

10 balls are distributed among three boxes. Probability that the first box will contain 3 balls is (a) \(\left[\left({ }^{10} \mathrm{C}_{3} \times 2^{7}\right) / 3^{10}\right]\) (b) \(\left[\left({ }^{10} \mathrm{C}_{3} \times 2^{7}\right) / 10^{3}\right]\) (c) \(\left[\left({ }^{10} \mathrm{C}_{3} \cdot{ }^{7} \mathrm{C}_{2}\right) / 3^{10}\right]\) (d) \(\left[\left({ }^{10} \mathrm{P}_{3} \cdot 2^{7}\right) / 3^{10}\right]\)

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