Problem 15 - Algebra for College Students

Question

A jar contains 10 balls numbered 1 through 10. Two balls are randomly selected one at a time without replacement. What is the probability that a) 1 is selected first and 2 is selected second? b) the sum of the numbers selected is 3? c) the sum of the numbers selected is 6?

Step-by-Step Solution

Answer

a) \( \frac{1}{90} \); b) \( \frac{1}{45} \); c) \( \frac{2}{45} \).

Step 1 - Understand the Problem

We have a jar with 10 balls numbered 1 through 10. Two balls are drawn one after the other without replacement. We need to find the probability for three different scenarios.

Step 2a - Find the probability for 1 being selected first and 2 selected second

Since we are not replacing the balls, the probability of drawing ball 1 first is \(\frac{1}{10}\). Once ball 1 is selected, there are 9 balls left. The probability of drawing ball 2 next is \(\frac{1}{9}\). Multiply these probabilities to get the overall probability: \[ P(a) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \].

Step 3a - Express the Probability

The probability of 1 being selected first and 2 being selected second is \(\frac{1}{90}\).

Step 2b - Find the probability that the sum of the numbers selected is 3

To get the sum of 3, the only pairs possible are (1, 2) and (2, 1). Since the balls are drawn without replacement, calculate the probability for each case and then sum them. \(P(1, 2) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90}\), and \(P(2, 1) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90}\).

Step 3b - Express the Combined Probability

These events are mutually exclusive, so add their probabilities: \[P(b) = \frac{1}{90} + \frac{1}{90} = \frac{2}{90} = \frac{1}{45} \].

Step 2c - Find the probability that the sum of the numbers selected is 6

To get a sum of 6, we have the pairs (1, 5), (2, 4), (4, 2), and (5, 1). Calculate the probabilities for each pair and sum them: \( P(1, 5) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \), \( P(2, 4) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \), \( P(4, 2) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \), \( P(5, 1) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \).

Step 3c - Express the Combined Probability

These events are also mutually exclusive, so add their probabilities: \[P(c) = \frac{1}{90} + \frac{1}{90} + \frac{1}{90} + \frac{1}{90} = \frac{4}{90} = \frac{2}{45} \].

Key Concepts

Probability Without Replacement

Imagine a jar with 10 balls numbered from 1 to 10. When we draw two balls one after the other without replacing the first ball before drawing the second, we are dealing with 'probability without replacement.' This means that the total number of balls decreases with each draw. The chance of drawing a specific ball changes after each pick because we don't put the ball back in the jar.

For example, if you draw the first ball and it is ball number 1, the probability of that happening is \(\frac{1}{10}\). Now, only 9 balls remain for the second draw. If we want to draw ball number 2 next, the probability is \(\frac{1}{9}\). The overall probability of drawing ball 1 first and then ball 2 is calculated by multiplying these probabilities: \[ P(1 \,and\, 2) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \]. This product represents the reduced likelihood due to the absence of replacements, thus altering the sample space after each draw.

Mutually Exclusive Events

Mutually exclusive events are scenarios where one event occurring means the other cannot. In our example, if you draw two balls and one is number 1 and the other is number 2, these two events (drawing 1 first, then 2 and vice versa) can't happen simultaneously. We handle these by adding up their individual probabilities.

For instance, to find the probability that the sum of the numbers drawn equals 3, we look at all possible pairs. There are two possible pairs for drawing sums of 3: (1, 2) and (2, 1). Since these events are mutually exclusive, we calculate the individual probabilities and add them up: \[ P(1, 2) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90}\, and \, P(2, 1) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \]. Summing these, we get: \[ P(sum \, of \, 3) = \frac{1}{90} + \frac{1}{90} = \frac{2}{90} = \frac{1}{45} \]. We sum probabilities because mutually exclusive events do not overlap.

Simple Probability Calculations

Simple probability involves calculating the likelihood of single events, often as fractions. Each possible event is considered, and their probabilities are combined for more complex cases. Consider drawing two balls, aiming for various sums. This requires calculating each pair's probability and summing them.

For example, finding the probability that the sum of drawn numbers is 6 involves possible pairs like (1, 5), (2, 4), (4, 2), and (5, 1). Calculate individual probabilities for each pair:

• \( P(1, 5) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \)
• \( P(2, 4) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \)
• \( P(4, 2) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \)
• \( P(5, 1) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \)

Summing these, we get: \[ P(sum \, of \, 6) = \frac{1}{90} + \frac{1}{90} + \frac{1}{90} + \frac{1}{90} = \frac{4}{90} = \frac{2}{45} \]. Simple probabilities are multiplied for each event and summed when events are mutually exclusive.