SOLUTION: 49% of all violent felons in the prison system are repeat offenders. If 35 violent felons are randomly selected, find the probability that a. Exactly 17 of them are repeat o

Question 1199758: 49% of all violent felons in the prison system are repeat offenders. If 35 violent felons are randomly selected, find the probability that

a. Exactly 17 of them are repeat offenders.
b. At most 17 of them are repeat offenders.
c. At least 16 of them are repeat offenders.
d. Between 15 and 21 (including 15 and 21) of them are repeat offenders.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**a) Probability of exactly 17 repeat offenders:**
* This is a binomial probability problem.
* n = 35 (sample size)
* p = 0.49 (probability of success - repeat offender)
* q = 1 - p = 0.51 (probability of failure - not a repeat offender)
* x = 17 (number of successes)
* P(X = 17) = (35C17) * (0.49)^17 * (0.51)^(35-17)
* where (35C17) is the number of combinations of 35 things taken 17 at a time.
* Using a binomial probability calculator or statistical software:
P(X = 17) ≈ 0.1223
**b) Probability of at most 17 repeat offenders:**
* P(X ≤ 17) = P(X = 0) + P(X = 1) + ... + P(X = 17)
* Use a cumulative binomial probability calculator or statistical software:
P(X ≤ 17) ≈ 0.6143
**c) Probability of at least 16 repeat offenders:**
* P(X ≥ 16) = P(X = 16) + P(X = 17) + ... + P(X = 35)
* Alternatively:
P(X ≥ 16) = 1 - P(X ≤ 15)
P(X ≥ 16) ≈ 1 - 0.4098 = 0.5902
**d) Probability of between 15 and 21 repeat offenders (inclusive):**
* P(15 ≤ X ≤ 21) = P(X = 15) + P(X = 16) + ... + P(X = 21)
* Use a cumulative binomial probability calculator or statistical software to find:
P(15 ≤ X ≤ 21) ≈ 0.7586
**In summary:**
* a) P(X = 17) ≈ 0.1223
* b) P(X ≤ 17) ≈ 0.6143
* c) P(X ≥ 16) ≈ 0.5902
* d) P(15 ≤ X ≤ 21) ≈ 0.7586

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