A certain virus afflicted the families in five adjacent houses in a row of 10 houses. If houses were randomly chosen from a row of 10 houses, what is the probability that the five houses would be adjacent? (Round your answer to three decimal places.)

We are tasked with finding the probability that five adjacent houses are chosen from a row of ten houses. Here's a step-by-step breakdown of how we can solve this:

### Total possible ways to choose 5 houses out of 10:
The total number of ways to choose 5 houses from 10 is given by the combination formula:

\[
\text{Total combinations} = \binom{10}{5}
\]

The combination formula is:

\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

Substituting \( n = 10 \) and \( r = 5 \):

\[
\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!}
\]

### Number of ways to choose 5 adjacent houses:
To choose 5 adjacent houses, we only need to pick a starting point for the group of 5 houses. The first house can start anywhere between the 1st and the 6th house, so there are 6 ways to choose a group of 5 adjacent houses.

### Probability:
The probability is the ratio of favorable outcomes (choosing 5 adjacent houses) to the total possible outcomes (choosing any 5 houses from 10):

\[
\text{Probability} = \frac{\text{Number of ways to choose 5 adjacent houses}}{\text{Total combinations}} = \frac{6}{\binom{10}{5}}
\]

Now, let's calculate the values.

\[
\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]

Therefore:

\[
\text{Probability} = \frac{6}{252} = \frac{1}{42} \approx 0.024
\]

### Final Answer:
The probability that the five houses are adjacent is approximately \( \boxed{0.024} \).

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. What is the probability of selecting 4 adjacent houses from a row of 10 houses?
2. How many different ways can you select 6 houses from 12, ensuring they are adjacent?
3. What is the probability of selecting 3 adjacent houses from a group of 8 houses?
4. In how many ways can you select 7 non-adjacent houses from a row of 15 houses?
5. How does the probability change if there are 20 houses in the row?

**Tip**: Understanding combinations and the relationship between favorable and total outcomes is key to calculating probabilities in such problems.

Explore the probability problem of choosing 5 adjacent houses from a row of 10. The solution includes a step-by-step breakdown using combinatorics and probability, along with key formulas such as the combination formula. Perfect for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation