(Please provide necessary R statements) 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?

To solve this problem, we are dealing with a **geometric distribution** because each call represents a Bernoulli trial (success = connecting the call, failure = busy signal) and we are interested in the probability of the first success occurring on the \(k\)-th trial. The probability of success on each call is 0.02, and the calls are independent.

In a geometric distribution, the probability mass function (PMF) for the first success on the \(k\)-th trial is given by:
\[
P(X = k) = (1 - p)^{k-1} \cdot p
\]
where \(p\) is the probability of success (0.02) and \(1 - p\) is the probability of failure (0.98).

We can use this to solve the three parts:

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

Here, \(k = 10\) and \(p = 0.02\), so we use the PMF formula:

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

In R, this can be computed as:

```R
# Probability of success
p <- 0.02

# Part (a) Probability that first success is on the 10th call
k <- 10
prob_a <- (1 - p)^(k - 1) * p
prob_a
```

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

The probability that it takes more than \(k\) calls to get the first success is given by:

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

In R, this can be computed as:

```R
# Part (b) Probability that it takes more than 5 calls to connect
k <- 5
prob_b <- (1 - p)^k
prob_b
```

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

For a geometric distribution, the mean (expected number of trials until the first success) is given by:

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

In R, this can be computed as:

```R
# Part (c) Mean number of calls to connect
mean_calls <- 1 / p
mean_calls
```

### Full R code:

```R
# Probability of success
p <- 0.02

# Part (a) Probability that first success is on the 10th call
k <- 10
prob_a <- (1 - p)^(k - 1) * p
prob_a

# Part (b) Probability that it takes more than 5 calls to connect
k <- 5
prob_b <- (1 - p)^k
prob_b

# Part (c) Mean number of calls to connect
mean_calls <- 1 / p
mean_calls
```

This R code will give you the results for all three parts.

---

Would you like further details on any part of the solution or clarification on the geometric distribution?

Here are some related questions:
1. What is the probability of connecting on the 1st call?
2. How does the probability change if the success rate increases to 0.05?
3. What is the probability of connecting after exactly 20 calls?
4. What is the variance in the number of calls needed to connect?
5. How does the expected number of calls change if the connection probability decreases?

**Tip:** In a geometric distribution, the probability of success does not change after each trial, making it memoryless. This feature simplifies many problems!

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?

Solve a geometric distribution problem where each radio station call has a 0.02 chance of connecting. Learn how to calculate the probability of connecting on the 10th call, the chance it takes more than 5 calls, and the mean number of calls needed. Suitable for college-level statistics.

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