n = 8 nights
p = 0.30 (probability of receiving a present)
You will need the formula for the binomial probability with X~Bin(n, p).
P(X = x) = C(n, x)*[px(1 - p)n-x]
A) P(X = 4) = C(8, 4)*[(0.30)4(0.70)4] = 0.1361
B) P(X ≥ 2) = C(8, 2)*[(0.30)2(0.70)6] + C(8, 3)*[(0.30)3(0.70)5] + ... + C(8, 8)*[(0.30)8(0.70)0] = 0.7447
You can also do P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.2553 = 0.7447.
C) This problem will require the conditional probability.
P(X < 3) = C(8, 0)*[(0.30)0(0.70)8] + C(8, 1)*[(0.30)1(0.70)7] + C(8, 2)*[(0.30)2(0.70)6] = 0.5518
P(X = 2) = C(8, 2)*[(0.30)2(0.70)6] = 0.2965
P(X = 2 | X < 3) = P(X = 2 and X < 3) / P(X < 3) = (0.2965)*(0.5518) / (0.5518) = 0.2965
The probability that Michael receives exactly 4 presents is 0.1361.
The probability that Michael receives at least 2 presents is 0.7447.
The probability that Michael receives 2 presents given that he receives less than 3 is 0.2965.
