The phone lines to an airline reservation system are occupied 40% of the time. Assume that the events that the lines are occupied on successive calls are independent. Assume that 10 calls are placed to the airline. (a) What is the probability that for exactly three calls, the lines are occupied? (b) What is the probability that for at least one call, the lines are not occupied? (c) What is the expected number of calls in which the lines are all occupied?

This problem can be tackled using binomial probability and the concept of independence for successive calls.

Let’s break it down into each part:

### Part (a): Probability that for exactly 3 calls, the lines are occupied

This is a **binomial probability problem** because we have a fixed number of trials (10 calls), two outcomes (occupied or not occupied), and the probability of success (occupied) is constant.

- The probability that the lines are occupied on a given call is \( p = 0.40 \).
- The number of calls is \( n = 10 \).
- We are interested in exactly \( k = 3 \) occupied calls.

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 \( \frac{n!}{k!(n-k)!} \).
- \( p^k \) is the probability of getting \( k \) successes.
- \( (1 - p)^{n - k} \) is the probability of getting \( n - k \) failures.

Substitute the values:

\[
P(X = 3) = \binom{10}{3} (0.40)^3 (0.60)^7
\]

### Part (b): Probability that for at least one call, the lines are **not** occupied

This is equivalent to calculating the complement of the probability that the lines are occupied for **all 10 calls**.

- The probability that the lines are not occupied on a given call is \( 1 - 0.40 = 0.60 \).
- The probability that the lines are occupied on all 10 calls is \( (0.40)^{10} \).

So the probability that at least one call is **not** occupied is:

\[
P(\text{at least one not occupied}) = 1 - (0.40)^{10}
\]

### Part (c): The expected number of calls in which the lines are all occupied

The expected value \( E(X) \) of a binomial distribution is given by:

\[
E(X) = n \cdot p
\]

Where:
- \( n = 10 \) is the number of trials.
- \( p = 0.40 \) is the probability of success (lines are occupied).

Thus:

\[
E(X) = 10 \times 0.40
\]

Let me compute the exact values for these expressions.Here are the results for each part:

### Part (a): 
The probability that exactly 3 calls out of 10 are occupied is approximately \( 0.215 \) (or 21.5%).

### Part (b): 
The probability that for at least one call, the lines are not occupied is approximately \( 0.9999 \) (or 99.99%).

### Part (c): 
The expected number of calls in which the lines are occupied is \( 4 \).

Would you like further details on any of these, or have any additional questions?

Here are 5 related questions for practice:
1. What is the probability that all 10 calls are occupied?
2. What is the probability that fewer than 5 calls are occupied?
3. What is the variance of the number of occupied calls?
4. What is the probability that between 2 and 5 calls are occupied?
5. How would the expected value change if the probability of occupation increased to 50%?

**Tip:** The expected value in a binomial distribution is simply \( n \times p \), so you can quickly determine how changes in \( p \) or \( n \) affect expectations.

Solve a binomial probability problem involving an airline reservation system where the phone lines are occupied 40% of the time. Calculate the probability of exactly 3 occupied calls, at least one non-occupied call, and the expected number of occupied calls. Suitable for Grades 10-12.

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