SOLUTION: 60% of all violent felons in the prison system are repeat offenders. If 45 violent felons are randomly selected, find the probability that a. Exactly 26 of them are repeat o

Here's how to solve this binomial probability problem:

* **n** (number of trials) = 45
* **p** (probability of success - repeat offender) = 0.60
* **q** (probability of failure - not a repeat offender) = 1 - p = 0.40

The binomial probability formula is: P(x) = (nCx) * p^x * q^(n-x)

Where nCx represents "n choose x" (the binomial coefficient).

**(a) Exactly 26 are repeat offenders:**

P(x = 26) = (45C26) * (0.60)^26 * (0.40)^19
P(x = 26) ≈ 0.1143

**(b) At most 29 are repeat offenders:**

This means 0 to 29 are repeat offenders. This is a cumulative probability; we need P(x ≤ 29).  
It's best to use a binomial cumulative distribution function (CDF) calculator or statistical software for this.

P(x ≤ 29) ≈ 0.7751

**(c) At least 29 are repeat offenders:**

This means 29 to 45 are repeat offenders. We can use the complement rule:

P(x ≥ 29) = 1 - P(x < 29) = 1 - P(x ≤ 28)

Use a binomial CDF calculator:
P(x ≥ 29) = 1 - 0.6728
P(x ≥ 29) ≈ 0.3272

**(d) Between 25 and 33 (inclusive):**

This means 25, 26, 27, 28, 29, 30, 31, 32, and 33 are repeat offenders. We can use the CDF:

P(25 ≤ x ≤ 33) = P(x ≤ 33) - P(x ≤ 24)

Use a binomial CDF calculator:
P(25 ≤ x ≤ 33) = 0.9784 - 0.2223
P(25 ≤ x ≤ 33) ≈ 0.7561

**Summary of Answers:**

* (a) P(x = 26) ≈ 0.1143
* (b) P(x ≤ 29) ≈ 0.7751
* (c) P(x ≥ 29) ≈ 0.3272
* (d) P(25 ≤ x ≤ 33) ≈ 0.7561