49% of all violent felons in the prison system are repeat offenders. If 45 violent felons are randomly selected, find the probability that a. Exactly 24 of them are repeat offenders. b. At most 23 of them are repeat offenders. c. At least 23 of them are repeat offenders. d. Between 15 and 20 (including 15 and 20) of them are repeat offenders.

0.100 ;

0.6673 ;

0.446 ;

0.311

Explanation:

Given that :

p = 49% = 0.49

Sample size, n = 45

From binomial probability formula:

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Exactly 24 of them are repeat offenders.

P(x = 24) = 45C24 * 0.49^24 * 0.51^21 = 0.100

b. At most 23 of them are repeat offenders.

P(x ≤ 23) = p(x = 0) + p(x = 1)... + p(x =23)

P(x ≤ 23) = 0.6673 (using online binomial probability calculator)

c. At least 23 of them are repeat offenders.

P(x ≥ 23) = p(x = 23) + p(x = 24)... + p(x =45)

P(x ≥ 23) = 0.445 (using online binomial probability calculator)

d. Between 15 and 20 (including 15 and 20) of them are repeat offenders.

P(x = 15) + p(x = 16) + p(x = 17) p(x = 18) p(x = 19) +

p(x = 20)

0.0131 + 0.0236 + 0.0387 + 0.0579 + 0.0790 + 0.0987

= 0.311