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Chapter 35: Problem 1109

An urn contains 6 white, 4 black, and 2 red balls. In a single draw, find the probability of drawing: (a) a red ball; (b) a black ball; (c) either a white or a black ball. Assume all outcomes equally likely.

Short Answer

Expert verified
The probabilities are: (a) The probability of drawing a red ball is \( \frac{1}{6} \). (b) The probability of drawing a black ball is \( \frac{1}{3} \). (c) The probability of drawing either a white or a black ball is \( \frac{5}{6} \).

Step by step solution

01

Calculate the probabilities

To find the probability of each scenario, we'll use the following formula: P(E) = (number of favorable outcomes) / (total number of possible outcomes)
02

Step 1(a): Probability of drawing a red ball

We want to find the probability of drawing a red ball (2 red balls are present). There are 12 balls in total, so the probability is: P(Red) = \( \frac{2}{12} = \frac{1}{6} \)
03

Step 1(b): Probability of drawing a black ball

We want to find the probability of drawing a black ball (4 black balls are present). There are 12 balls in total, so the probability is: P(Black) = \( \frac{4}{12} = \frac{1}{3} \)
04

Step 1(c): Probability of drawing either a white or a black ball

We want to find the probability of drawing either a white or a black ball (6 white balls and 4 black balls are present). There are 12 balls in total, so the probability is: P(White or Black) = \( \frac{6+4}{12} = \frac{10}{12} = \frac{5}{6} \) So the probabilities are: (a) The probability of drawing a red ball is \( \frac{1}{6} \). (b) The probability of drawing black ball is \( \frac{1}{3} \). (c) The probability of drawing either a white or a black ball is \( \frac{5}{6} \).

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