Among 60 automobile repair parts loaded on a truck, 45 are destined for Seattle and 15 for Vancouver. If two of the parts are unloaded in Portland by mistake. What are the probabilities that (a) both parts should have gone to Seattle
(b) Both parts should have gone to Vancouver

© one should have gone to Seattle and one to Vancouver?

We are asked to find the probabilities that two mistakenly unloaded parts should have gone to Seattle, Vancouver, or one to each location. Here's how we can calculate them:

### Step 1: Total Combinations
First, let's calculate the total number of ways to choose 2 parts from the 60 available parts. This is a combination problem:
\[
\text{Total combinations} = \binom{60}{2} = \frac{60 \times 59}{2} = 1770
\]

### Step 2: Probabilities for Each Case
Now, we calculate the probabilities for each case.

#### (a) Probability that both parts should have gone to Seattle

There are 45 parts destined for Seattle. The number of ways to choose 2 parts from the 45 Seattle parts is:
\[
\binom{45}{2} = \frac{45 \times 44}{2} = 990
\]
Thus, the probability is:
\[
P(\text{both to Seattle}) = \frac{\binom{45}{2}}{\binom{60}{2}} = \frac{990}{1770} \approx 0.559
\]

#### (b) Probability that both parts should have gone to Vancouver

There are 15 parts destined for Vancouver. The number of ways to choose 2 parts from the 15 Vancouver parts is:
\[
\binom{15}{2} = \frac{15 \times 14}{2} = 105
\]
Thus, the probability is:
\[
P(\text{both to Vancouver}) = \frac{\binom{15}{2}}{\binom{60}{2}} = \frac{105}{1770} \approx 0.0593
\]

#### (c) Probability that one part should have gone to Seattle and one to Vancouver

To calculate this, we find the number of ways to choose 1 part from the 45 Seattle parts and 1 part from the 15 Vancouver parts:
\[
\binom{45}{1} \times \binom{15}{1} = 45 \times 15 = 675
\]
Thus, the probability is:
\[
P(\text{one to Seattle, one to Vancouver}) = \frac{45 \times 15}{\binom{60}{2}} = \frac{675}{1770} \approx 0.381
\]

### Final Probabilities Summary
1. \( P(\text{both to Seattle}) \approx 0.559 \)
2. \( P(\text{both to Vancouver}) \approx 0.0593 \)
3. \( P(\text{one to Seattle, one to Vancouver}) \approx 0.381 \)

Would you like a detailed breakdown of the steps or further clarification?

Here are 5 related questions:
1. How do you calculate combinations for more than two selections?
2. What happens to probabilities if three parts are unloaded by mistake?
3. Can probabilities change if more destinations are added (e.g., a third city)?
4. What is the complementary probability of none of the parts going to Seattle?
5. How would you approach this problem with dependent events?

**Tip:** When dealing with combination problems, always remember that order does not matter!

Among 60 automobile repair parts loaded on a truck, 45 are destined for Seattle and 15 for Vancouver. If two of the parts are unloaded in Portland by mistake. What are the probabilities that (a) both parts should have gone to Seattle (b) both parts should have gone to Vancouver (c) one should have gone to Seattle and one to Vancouver?

Calculate the probability of two mistakenly unloaded parts from a truck with 60 automobile repair parts going to either Seattle or Vancouver. The solution involves combinatorics and basic probability, providing detailed steps for each case. Ideal for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation