Thirty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty? (Round your answer to three decimal places.)

The problem at hand involves calculating the probability that exactly 3 out of 10 phones purchased by a company will be replaced under warranty.

Let’s break this down:

- 30% of all telephones are submitted for service under warranty.
- Of these phones submitted, 30% are replaced (while the remaining 70% are repaired).
  
### Step 1: Find the overall probability that a phone will be replaced
The probability that a phone is submitted for service and replaced is the product of these two probabilities:
\[
P(\text{replaced}) = P(\text{submitted}) \times P(\text{replaced given submitted}) = 0.30 \times 0.30 = 0.09.
\]
Thus, the probability that any given phone will be replaced is \(0.09\).

### Step 2: Binomial Probability Calculation
We need to calculate the probability that exactly 3 out of 10 phones will be replaced. This is a binomial probability problem, where:
- \(n = 10\) (the number of phones),
- \(k = 3\) (the number of phones to be replaced),
- \(p = 0.09\) (the probability of a phone being replaced).

The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k},
\]
where \(\binom{n}{k}\) is the binomial coefficient, calculated as:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}.
\]

### Step 3: Apply the Formula
Substitute the values \(n = 10\), \(k = 3\), and \(p = 0.09\) into the formula:

\[
P(X = 3) = \binom{10}{3} (0.09)^3 (1 - 0.09)^{7}.
\]

Let me compute this.The probability that exactly 3 out of 10 phones will be replaced under warranty is \(0.045\), rounded to three decimal places.

Would you like more details or have any questions?

Here are some related questions to further explore the topic:
1. What is the probability that no phones will be replaced under warranty?
2. What is the expected number of phones that will be replaced?
3. What is the variance of the number of phones that will be replaced?
4. What is the probability that at most 2 phones will be replaced under warranty?
5. How does the probability change if the percentage of phones replaced increases to 40%?

**Tip**: The binomial distribution is ideal for calculating probabilities in situations where there are only two outcomes (success or failure) across multiple independent trials.

Thirty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty? (Round your answer to three decimal places.)

Calculate the probability that exactly 3 out of 10 phones will be replaced under warranty. This problem explores key probability concepts, using binomial distribution and the binomial theorem. Suitable for students in Grades 11-12.

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