Using the formula for the multinomial distribution, the probability of when n = 5, = 0.1 is
A){5!/(2! × 1! × 2!)} ×
B)2! × 1! × 2!
D){5!/(0.4! × 0.5! × 0.1!)} ×

Question

Using the formula for the multinomial distribution, the probability of  when n = 5,  = 0.1 is
A){5!/(2! × 1! × 2!)} × 
B)2! × 1! × 2! 
D){5!/(0.4! × 0.5! × 0.1!)} ×

Quizplus · Accepted Answer

The formula for a multinomial distribution with n trials and probabilities p1, p2, ..., pk for k possible outcomes is:
P(X1 = x1, X2 = x2, ..., Xk = xk) = n! / (x1! x2! ... xk!) * (p1^x1) * (p2^x2) * ... * (pk^xk)
In this case, there are 5 trials and 3 possible outcomes (e, b, and TB8376_11). The probabilities are p(e) = 0.1, p(b) = 0.4, p(TB8376_11) = 0.5. We want to find the probability of getting 1 e, 2 b's, and 2 TB8376_11's.
Plugging in the values, we get:
P(Xe = 1, Xb = 2, XTB8376_11 = 2) = 5! / (1! 2! 2!) * (0.1^1) * (0.4^2) * (0.5^2)
Simplifying:
P(Xe = 1, Xb = 2, XTB8376_11 = 2) = (5! / (2! 1! 2!)) * 0.01 * 0.16 * 0.25
P(Xe = 1, Xb = 2, XTB8376_11 = 2) = 0.015
Therefore, the answer is A. The formula for the multinomial distribution is:
P(X1 = r1, X2 = r2, ..., Xk = rk) = n!/(r1!r2!...rk!) * p1^r1 * p2^r2 * ... * pk^rk
where X1, X2, ..., Xk are the k categories, r1, r2, ..., rk are the corresponding number of successes, n is the total number of trials, and p1, p2, ..., pk are the corresponding probabilities of successes.
In this scenario, there are 5 trials (n = 5) and 3 categories (k = 3), with probabilities of success 0.1, 0.4, and 0.5.
We want to find the probability of getting 2 successes in the first category, 1 success in the second category, and 2 successes in the third category:
P(X1 = 2, X2 = 1, X3 = 2) = 5!/(2!1!2!) * (0.1)^2 * (0.4)^1 * (0.5)^2
Simplifying:
P(X1 = 2, X2 = 1, X3 = 2) = (5x4x3x2x1)/(2x1x1x2x1x2) * 0.01 * 0.4 * 0.25
P(X1 = 2, X2 = 1, X3 = 2) = 0.15
Therefore, the best choice is A.

Using the Formula for the Multinomial Distribution, the Probability of