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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

Standard deviation of four consecutive numbers which are in arithmetic series is \(\sqrt{5}\) then common difference of this series is (a) \(\sqrt{5}\) (b) \(2 \cdot \sqrt{5}\) (c) \(\pm 2\) (d) \(\sqrt{2}\)

The median of first \(\mathrm{n}+3\) natural number is (a) \([(n+4) / 2]\) (b) \([(n-4) / 2]\) (c) \([(n-1) / 2]\) (d) \([(n+3) / 4]\)

The mean of five observations is \(4.0\) and variance is \(5.2\) among five three observations are \(1,2,6\) then remaining observation are (a) 5,10 (b) 4,7 (c) 3,10 (d) 5,8

If the mean deviation of the number \(1,1+\mathrm{d}, 1+2 \mathrm{~d}, \ldots \ldots 1+50 \mathrm{~d}\). From their mean is 260 then d is (a) \(20.5\) (b) \(20.3\) (c) \(20.4\) (d) \(10.4\)

A \(2 \times 2\) determinant is such that all its entries are \(1,-1\) or \(0 .\) If one determinant is chosen from such determinants what is the probability that the value of the determinant is zero? (a) \((3 / 8)\) (b) \((11 / 27)\) (c) \((2 / 9)\) (d) \((25 / 81)\)
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