A six sided unbiased die with four green faces and two red faces is rolled…

Key idea: Use the binomial distribution with n = 7 trials and probability of green p = 4/6 = 2/3.

• Step 1: Probability of exactly k green faces is P(k) = C(7,k)*(2/3)^k*(1/3)^(7-k).

• Step 2: The mode of a binomial distribution is floor((n+1)p). Here (n+1)p = 8*(2/3) = 16/3, so the mode is floor(16/3) = 5. Thus five green faces is the most likely count.

• Step 3: Compare the probabilities for the outcomes in the choices (using 3^7 = 2187 so (2/3)^k*(1/3)^(7-k) = 2^k / 2187):

  • Three green, four red: P = C(7,3)*2^3/2187 = 35*8/2187 = 280/2187 ≈ 0.128.

  • Four green, three red: P = C(7,4)*2^4/2187 = 35*16/2187 = 560/2187 ≈ 0.256.

  • Five green, two red: P = C(7,5)*2^5/2187 = 21*32/2187 = 672/2187 ≈ 0.307.

  • Six green, one red: P = C(7,6)*2^6/2187 = 7*64/2187 = 448/2187 ≈ 0.205.

Conclusion: Five green faces and two red faces is the most likely outcome, with probability 672/2187 ≈ 0.307.