The image contains a probability problem involving a scenario where Cade is filling out summer job applications. Here's a detailed breakdown:

Problem:

1. Cade is submitting 5 applications.
2. The probability that each of his applications gets a job offer is 61% (or 0.61).
3. You are asked to calculate the probability that 3 out of the 5 applications result in job offers.
4. The result should be given as a decimal rounded to four decimal places.

Solution Approach:

This is a binomial probability problem where:

• $n = 5$ (the number of trials, i.e., applications submitted)
• $k = 3$ (the number of successes, i.e., job offers)
• $p = 0.61$ (the probability of success for each trial, i.e., getting a job offer)
• $q = 1 - p = 0.39$ (the probability of failure for each trial, i.e., not getting a job offer)

The binomial probability formula is:

$P(k) = \binom{n}{k} p^k (1 - p)^{n-k}$

Where:

• $\binom{n}{k}$ is the binomial coefficient calculated as:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

I will now compute this probability.The probability that Cade will receive job offers for exactly 3 out of the 5 applications is 0.3452, rounded to four decimal places.

Would you like any more details or have any further questions?

Here are 5 related questions you might explore:

1. How do you calculate the probability of receiving offers for exactly 4 out of 5 applications?
2. What is the probability that Cade receives at least one job offer from his 5 applications?
3. How does the probability change if the chance of getting a job offer increases to 70%?
4. What if Cade submits 7 applications instead of 5? How does this affect the probability for 3 offers?
5. How do you calculate the expected number of job offers Cade will receive?

Tip: In binomial probability problems, it's useful to identify whether you need the probability for "exactly," "at least," or "at most" a certain number of successes. These nuances affect the approach!