Suppose you just received a shipment of fifteen televisions. Four of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?

To solve this, we can break the problem into two parts. First, we compute the probability that both televisions work, and then we calculate the probability that at least one of the televisions does not work.

### Part 1: Probability that both televisions work
There are 15 televisions in total, and 4 are defective. This means 11 of the televisions work. We need to calculate the probability that both selected televisions are not defective (i.e., they both work).

We will use **combinations** to solve this. The total number of ways to select 2 televisions from 15 is given by:

\[
\binom{15}{2} = \frac{15 \times 14}{2 \times 1} = 105
\]

The number of ways to select 2 working televisions from the 11 that work is:

\[
\binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55
\]

Thus, the probability that both selected televisions work is:

\[
P(\text{both work}) = \frac{\binom{11}{2}}{\binom{15}{2}} = \frac{55}{105} = \frac{11}{21}
\]

### Part 2: Probability that at least one of the two televisions does not work
The probability that at least one of the selected televisions does not work is the complement of the probability that both televisions work. In other words:

\[
P(\text{at least one does not work}) = 1 - P(\text{both work})
\]

Using the result from part 1:

\[
P(\text{at least one does not work}) = 1 - \frac{11}{21} = \frac{10}{21}
\]

### Final Answers:
1. The probability that both televisions work is \( \frac{11}{21} \).
2. The probability that at least one of the two televisions does not work is \( \frac{10}{21} \).

---

Would you like more details on any part of the solution? 

Here are five questions related to the problem:
1. How do you calculate combinations and why are they used in this problem?
2. What would the probability be if only one defective television were in the shipment?
3. How does the probability change if more televisions are randomly selected?
4. What is the difference between dependent and independent events in probability?
5. Can you extend this approach to calculate probabilities for more complex systems?

**Tip**: For problems involving "at least one" outcomes, it is often easier to calculate the complement (opposite event) and subtract it from 1.

Discover how to calculate the probability that both selected televisions work in a shipment of 15 televisions, 4 of which are defective. Learn key probability and combinatorics concepts using the combination formula and complement rule. Suitable for 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