Question

On a game show, a contestant must choose one of four boxes to open. One box contains points towards the grand prize; the others are empty. There are 12 rounds in the game and in each round four new boxes are presented to the contestant. If they guess for every round, find the probability of the following events:

a. 7 correct games

b. 10 incorrect guesses

c. fewer than 5 correct guesses

d. more than 10 correct guesses

e. between 7 and 10 correct guesses (inclusive)

Answer 1

n = 12

p(correct) = 1/4 = 0.25

P(incorrect) = 1 - 0.25 = 0.75

It is a binomial distribution.

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

a) P(X = 7) = 12C7 * (0.25)^7 * (0.75)^0 = 0.0483

b) P(X = 10) = 12C10 * (0.75)^10 * (0.25)^2 = 0.2323

c) P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

                = 12C0 * (0.25)^0 * (0.75)^12 + 12C1 * (0.25)^1 * (0.75)^11 + 12C2 * (0.25)^2 * (0.75)^10 + 12C3 * (0.25)^3 * (0.75)^9 + 12C4 * (0.25)^4 * (0.75)^8 = 0.8424

d) P(X > 10) = P(X = 11) + P(X = 12)

                  = 12C11 * (0.25)^11 * (0.75)^1 + 12C12 * (0.25)^12 * (0.75)^0 = 0.0000

e) P(7 < X < 10) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

                        = 12C7 * (0.25)^7 * (0.75)^5 + 12C8 * (0.25)^8 * (0.75)^4 + 12C9 * (0.25)^9 * (0.75)^3 + 12C10 * (0.25)^10 * (0.75)^2 = 0.0143