Use the multinomial formula to find the probability for the situation in which n = 5, X₁ = 2, X₂ = 2, X₃ = 1, and, p₁ = 0.4, p₂ = 0.4, p₃ = 0.2.

Question

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The multinomial formula is given by: P(X1 = x1, X2 = x2, ..., Xk = xk) = n! / (x1! x2! ... xk!) * p1^x1 * p2^x2 * ... * pk^xk where n is the total number of trials, k is the number of possible outcomes, xi is the number of trials that result in outcome i, and pi is the probability of outcome i. In this case, we have k = 3 (since there are three possible outcomes: ₁, ₂, and ₃), n = 5, x1 = 2, x2 = 2, and x3 = 1. We are also given the probabilities p1 = 0.4, p2 = 0.4, and p3 = 0.2. Substituting these values into the multinomial formula, we get: P(X₁ = 2, X₂ = 2, X₃ = 1) = 5! / (2! 2! 1!) * 0.4^2 * 0.4^2 * 0.2^1 = 0.1536 Therefore, the answer is choice C.

Use the Multinomial Formula to Find the Probability for the Situation