 Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal. Assume that your calls are independent. (a) What is the probability that your first call that connects is your 10th call? (b) What is the probability that it requires more than five calls for you to connect? (c) What is the mean number of calls needed to connect?

Let's break down the problem step by step:

We are dealing with a **geometric distribution**, where:

- Each trial (call) is independent.
- The probability of success (getting connected) on any single call is \( p = 0.02 \).
- The probability of failure (getting a busy signal) on any single call is \( 1 - p = 0.98 \).

### (a) What is the probability that your first call that connects is your 10th call?

For this part, we need the probability that you have 9 failures followed by 1 success on the 10th call. The formula for the probability of the first success on the \(n\)-th trial (geometric distribution) is:

\[
P(X = n) = (1 - p)^{n-1} \cdot p
\]

Substituting the values \( p = 0.02 \) and \( n = 10 \):

\[
P(X = 10) = (0.98)^9 \cdot 0.02
\]

Let's calculate this.

The probability that your first call that connects is on your 10th call is approximately \( 0.0167 \) or 1.67%.

### (b) What is the probability that it requires more than five calls for you to connect?

The probability that it requires more than 5 calls to connect is the complement of the event that you connect in the first 5 calls. In other words, we need to find:

\[
P(X > 5) = 1 - P(X \leq 5)
\]

Now, \( P(X \leq 5) \) is the probability that you connect in the first 5 calls, which is the sum of the probabilities of connecting on the 1st, 2nd, 3rd, 4th, or 5th call. This can be computed as:

\[
P(X \leq 5) = 1 - (1 - p)^5
\]

Let’s calculate \( P(X > 5) \).The probability that it requires more than five calls for you to connect is approximately \( 0.904 \) or 90.4%.

### (c) What is the mean number of calls needed to connect?

For a geometric distribution, the mean (expected value) is given by the formula:

\[
E(X) = \frac{1}{p}
\]

Substituting \( p = 0.02 \):

\[
E(X) = \frac{1}{0.02}
\]

Let’s calculate the mean number of calls.The mean number of calls needed to connect is 50.

### Summary:
- (a) The probability that your first call that connects is your 10th call is approximately \(0.0167\) (1.67%).
- (b) The probability that it requires more than five calls to connect is approximately \(0.904\) (90.4%).
- (c) The mean number of calls needed to connect is 50.

Would you like more details on any of these calculations or have any further questions?

Here are five related questions to expand on these topics:
1. What is the probability that you connect on the first call?
2. What is the standard deviation of the number of calls needed to connect?
3. How would the probability change if the success rate was 5% instead of 2%?
4. What is the cumulative probability of connecting within 20 calls?
5. How would you simulate this scenario in a program to estimate the probability?

**Tip:** For geometric distributions, the expected number of trials to get the first success is always \( \frac{1}{p} \), where \( p \) is the probability of success.

Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal. Assume that your calls are independent. (a) What is the probability that your first call that connects is your 10th call? (b) What is the probability that it requires more than five calls for you to connect? (c) What is the mean number of calls needed to connect?

Explore a probability problem involving a radio station call-in scenario. Learn how to calculate the probability of connecting on the 10th call, the chances of needing more than five calls, and the mean number of calls needed. Understand the key concepts of geometric distribution and expected value. Suitable for advanced high school or college students.

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