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Chapter 3: Problem 5

An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black?

Short Answer

Expert verified
The probability of selecting 2 white balls first and then 2 black balls from an urn containing 6 white and 9 black balls is \(\frac{4}{9}\).

Step by step solution

01

Calculate Number of Favorable Outcomes

To find the number of favorable outcomes, we need to determine the total ways of selecting 2 white balls first and then 2 black balls. Since there are 6 white and 9 black balls, the number of ways to select 2 white balls first and 2 black balls later can be calculated using the multiplication principle: Number of ways to pick 2 white balls: \(\binom{6}{2} = 15\) Number of ways to pick 2 black balls: \(\binom{9}{2} = 36\) Number of favorable outcomes: \(15 \times 36 = 540\)
02

Calculate Total Possible Outcomes

Now, we need to find the total possible outcomes when selecting 4 balls from the urn without replacement. Since there are 15 balls in the urn, we can use combinations to determine the total possible ways to pick any 4 balls: Total possible outcomes: \(\binom{15}{4} = 1365\)
03

Compute the Probability

To find the probability of choosing 2 white balls first and then 2 black balls, we can divide the number of favorable outcomes by the total possible outcomes: Probability = \(\frac{favorable\:outcomes}{total\:possible\:outcomes}\) Probability = \(\frac{540}{1365}\)
04

Simplify the Probability

Finally, we can simplify the probability fraction by finding the greatest common divisor (GCD) between the numerator and denominator: GCD(540, 1365) = 135 Therefore, the simplified probability is \(\frac{540}{1365}\div\frac{135}{135}=\frac{4}{9}\). So the probability of selecting 2 white balls first and then 2 black balls is \(\frac{4}{9}\).

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