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Chapter 2: Problem 130

A box contains 4 black marbles, 3 red marbles, and 2 white marbles. What is the probability that a black marble, then a red marble, then a white marble is drawn without replacement?

Short Answer

Expert verified
The probability of drawing a black marble, a red marble, and a white marble consecutively without replacement is \(\frac{1}{21}\).

Step by step solution

01

1. Calculate the probability of drawing a black marble

There are 4 black marbles and a total of 9 marbles. Therefore, the probability of drawing a black marble first is: \[P(B) = \frac{number\: of\: black\: marbles}{total\: number\: of\: marbles} = \frac{4}{9}\]
02

2. Remove a black marble from the box and calculate the probability of drawing a red marble

After removing a black marble, there are 3 red marbles and 8 marbles total left in the box. The probability of drawing a red marble next is: \[P(R|B) = \frac{number\: of\: red\: marbles}{total\: number\: of\: marbles\: after\: drawing\: a\: black} = \frac{3}{8}\]
03

3. Remove a red marble from the box and calculate the probability of drawing a white marble

Now the box has 2 white marbles and 7 marbles total left in the box. The probability of drawing a white marble as the third marble is: \[P(W|R,B) = \frac{number\: of\: white\: marbles}{total\: number\: of\: marbles\: after\: drawing\: a\: black\: and\: a\: red} = \frac{2}{7}\]
04

4. Calculate the probability of drawing a black marble, a red marble, and a white marble consecutively

We multiply the probabilities of each step together to find the final probability: \[P(B, R, W) = P(B) \times P(R|B) \times P(W|R,B) = \frac{4}{9} \times \frac{3}{8} \times \frac{2}{7}\] \[P(B, R, W) = \frac{24}{504}\]
05

5. Simplify the final probability

We can simplify the final probability by dividing both the numerator and the denominator by the greatest common divisor (which is 24): \[P(B, R, W) = \frac{24}{504} = \frac{1}{21}\] The probability of drawing a black marble, a red marble, and a white marble consecutively without replacement is \(\frac{1}{21}\).

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