a. at r = 2
P = [10! / (10 – 2)! 2!] * 0.33^2 * 0.67^(10 – 2)
P = 0.1989 ~ 0.20 (answer)
b. We deduct the values of P when r = 0, r = 1, r = 2 from 1 to get all values when r > 2
at r = 0
P = [10! / (10 – 0)! 0!] * 0.33^0 * 0.67^(10 – 0)
P = 0.0182
at r = 1
P = [10! / (10 – 1)! 1!] * 0.33^1 * 0.67^(10 – 1)
P = 0.0898
so for r > 2, the probability is:
P = 1 – (0.1989 + 0.0182 + 0.0898)
P = 0.6931 (answer)
c. at r = 3
P = [10! / (10 – 3)! 3!] * 0.33^3 * 0.67^(10 – 3)
P = 0.2614
at r = 4
P = [10! / (10 – 4)! 4!] * 0.33^4 * 0.67^(10 – 4)
P = 0.2253
at r = 5
P = [10! / (10 – 5)! 5!] * 0.33^5 * 0.67^(10 – 5)
P = 0.1332
So when r is between 2 and 5 inclusive, the probability is:
P = 0.1989+ 0.2614 + 0.2253 + 0.1332
P = 0.8188 (answer)