Question

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

The probabilities of drawing different colored balls from the urn are: (a) red ball: \( \frac{1}{6} \), (b) black ball: \( \frac{1}{3} \), and (c) either a white or black ball: \( \frac{5}{6} \).

Step-by-step solution

Problem representation

Let's first represent the problem: - The urn contains 6 white, 4 black, and 2 red balls. - There are three questions: (a) probability of drawing a red ball, (b) probability of drawing a black ball, and (c) probability of drawing either a white or black ball. We need to find these probabilities.

Total number of outcomes

Calculate the total number of outcomes, which is the total number of balls in the urn. Total outcomes = Number of white balls + Number of black balls + Number of red balls = 6 + 4 + 2 = 12

Probability formula

We will be using the probability formula for all three cases: P(event) = number of successful outcomes / total number of outcomes

Probability of drawing a red ball (a)

Now, let's find the probability of drawing a red ball (event A): P(A) = Number of red balls / Total number of balls = 2 / 12 Simplify the fraction, and we get: P(A) = 1 / 6

Probability of drawing a black ball (b)

Next, let's find the probability of drawing a black ball (event B): P(B) = Number of black balls / Total number of balls = 4 / 12 Simplify the fraction: P(B) = 1 / 3

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

Finally, let's find the probability of drawing either a white or a black ball (event C): To find this probability, we can add the probabilities of drawing a white ball and drawing a black ball because these two events are mutually exclusive. P(C) = P(white ball) + P(black ball) First, calculate the probability of drawing a white ball: P(white ball) = Number of white balls / Total number of balls = 6 / 12 Simplify the fraction: P(white ball) = 1 / 2 Now, P(C) = P(white ball) + P(black ball) = (1 / 2) + (1 / 3) To add the fractions, find a common denominator (6): P(C) = (3 / 6) + (2 / 6) P(C) = 5 / 6 So, the probabilities are: (a) P(A = red ball) = 1 / 6 (b) P(B = black ball) = 1 / 3 (c) P(C = white or black ball) = 5 / 6

Key concepts

Probability of Drawing a Red Ball

To determine the probability of drawing a red ball from an urn, we need to follow some simple probability steps. The urn contains a total of 12 balls: 6 white, 4 black, and 2 red. Each outcome is equally likely.
The formula for finding probability is:

• \( P( ext{event}) = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} \)

For the red ball, our successful outcome is drawing one of the 2 red balls. Thus, the probability is:

• \( P( ext{red ball}) = \frac{2}{12} \)

When you simplify the fraction \( \frac{2}{12} \), you get \( \frac{1}{6} \).
So, the probability of drawing a red ball is a simple fraction of the favorable outcomes over the total possible draws, which calculates to \( \frac{1}{6} \). This tells us that if you were to draw a ball once, there is a one in six chance it will be red.

Probability of Drawing a Black Ball

Similarly, to calculate the probability of drawing a black ball, we use the same probability formula. In this urn, there are \( 4 \) black balls. So our desirable outcome here is to draw any one of these 4 black balls.
Applying our probability formula:

• \( P( ext{black ball}) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{4}{12} \)

Now we simplify \( \frac{4}{12} \), which equals \( \frac{1}{3} \).
This means that when you reach into the urn without looking, there is a one in three chance that you will grab a black ball. The odds here reflect the proportion of black balls relative to the total count of all the balls.

Mutually Exclusive Events

Mutually exclusive events are those that cannot happen at the same time. When considering probability, this concept helps to calculate the probability of either one event or another event happening.
In our urn example, drawing a white ball or drawing a black ball are mutually exclusive because a ball cannot be both white and black at the same time.

• For mutually exclusive events \( A \) and \( B \), the probability of either \( A \) or \( B \) occurring is the sum of their individual probabilities: \( P(A \cup B) = P(A) + P(B) \).

We calculated the probability of drawing a white ball as:- \( P( ext{white ball}) = \frac{6}{12} = \frac{1}{2} \).
Thus, to find the probability of either drawing a white or black ball:

• \( P( ext{white or black}) = P( ext{white ball}) + P( ext{black ball}) = \frac{1}{2} + \frac{1}{3} \)

Finding a common denominator (6), we add:

• \( P( ext{white or black}) = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \)

This shows that there is a five in six chance of drawing either a white or a black ball, given they don't overlap or happen simultaneously.