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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 10 th 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?

Short Answer

Expert verified
(a) 0.0167 (b) 0.9039 (c) 50

Step by step solution

01

Identify the Distribution

This problem involves a fixed probability of success (connecting a call) and seeks the number of trials until the first success. This scenario matches a geometric distribution, where each call is independent, and the probability of success (connecting a call) is 0.02.
02

Determine Parameters

For a geometric distribution, the probability of success on each trial is denoted as \( p = 0.02 \). The number of trials until the first success is \( X \), which follows a geometric distribution.
03

Solve Part (a)

To find the probability that the first call that connects is the 10th call, we use the formula for the geometric distribution: \( P(X=k) = (1-p)^{k-1} \times p \). Substituting \( k = 10 \) and \( p = 0.02 \), we get: \[ P(X=10) = (1-0.02)^{9} \times 0.02 \] \[ = 0.98^{9} \times 0.02 \] \[ \approx 0.0167 \].
04

Solve Part (b)

For finding the probability of requiring more than 5 calls to connect, we calculate the probability of failing the first 5 calls: \[ P(X > 5) = (1-p)^5 \] Substituting \( p = 0.02 \): \[ P(X > 5) = 0.98^5 \approx 0.9039 \].
05

Solve Part (c)

The mean (expected value) of a geometric distribution is given by \( \frac{1}{p} \). For this problem, the mean number of calls required to get a connection is: \[ E(X) = \frac{1}{0.02} = 50 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Success
In the context of a geometric distribution, the "probability of success" (denoted by the symbol \( p \)) is the likelihood that a single trial results in a desired outcome. Here, the desired outcome is successfully connecting a call. In our scenario, each time you attempt to call the radio station, there's a 0.02 chance you will connect without getting a busy signal.

This probability remains constant regardless of past attempts.
  • Probability of connecting on a given call: \( p = 0.02 \)
The probability of failure in a single call would then be \( 1-p \), or 0.98.

Understanding the probability of success is crucial as it's a fundamental parameter in computing other probabilities and the expected value within the geometric distribution framework.
Independent Trials
Independent trials refer to the concept that each attempt to achieve a success (connecting on the call) does not affect other attempts. Each call you make to the radio station plays out independently of previous or subsequent calls. That means the probability of getting through on any one call is always the same, unaffected by past successes or failures.

This independence is integral to the geometric distribution, ensuring that the sequence or history of outcomes doesn't alter the odds for each individual trial.

In mathematical terms, this implies that the events are statistically independent. Such assumptions allow us to employ probability models like the geometric distribution efficiently.
Expected Value
The expected value in a geometric distribution tells us the average number of trials needed to achieve the first success. It's a statistical mean, offering an insight into what to expect over a long sequence of trials.
  • The formula for calculating the expected value \( E(X) \) in a geometric distribution is \( \frac{1}{p} \).
For our problem, with a probability of success \( p = 0.02 \), the expected number of calls needed to connect is calculated as:\[E(X) = \frac{1}{0.02} = 50\]

This doesn't imply you'll always connect on the 50th call, but across many attempts, you'll average 50 calls to connect once.
Probability Distribution
Probability distribution in the context of geometric distribution provides a model for the probabilities associated with the number of trials needed for the first success. It can be visualized as a list or a graph showing all possible outcomes and their associated probabilities.
  • For example, the probability that the first successful connection is made on the 10th call can be calculated using the formula: \( P(X=k) = (1-p)^{k-1} \times p \).
For this problem:\[P(X=10) = 0.98^{9} \times 0.02 \approx 0.0167\]

This indicates that there's a 1.67% chance of connecting on exactly the 10th call. The distribution is tail-heavy, with higher probabilities at the start (fewer calls) and rapidly decreasing as the number of trials increases.

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