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Statistics word problem, HELP?

Participants in the TV show are taken from a large number of candidates with approximately equal numbers of men and women. Among the 11 remaining participants, there were only two women. If participants are selected at random, what is the likely to ensure that women 2 or less, where 11 people are taken?

By implementing a statistical test or trial, each with two possible outcomes, one the results can be arbitrarily assigned a "success" and the other a "failure." If the results of the first experiments have no effect on More recently, and the probability of "success" remains the same for all experiments, then the probability of getting exactly the positive results rn tries is given by the binomial probability. If p is the probability of any test with a positive result and q is the probability of failure, then the binomial probability P (n R) = Pr ^ * ^ * NCR q (nr), where NCR is the binomial coefficient given by n = nCr! / [R! * (No.)] While applicants are selected without replacement, group of potential candidates in this case is considered important enough for the effect of choosing a small sample of 11 people will have a negligible effect on the proportion males and females. In other words, statistics gathering test candidates are almost independent. Since there are roughly equal numbers of men and women, the chance to pick up a woman in any test, are considered a success, is about half. This is the value we attribute to P. Chance to choose a man is q, which is always 1 – p, will also be half. The probability of selection of candidates two or fewer 11 women surrounded by two women, a woman the female, each outcome global mutually exclusive of others. In other words, we can only a unique number of women in any one set of tests. We add these probabilities to obtain the probability of choosing no more than two women. This is P (2 women of 11 participants) = 11C2 * (1 / 2) ^ 2 * (1 / 2) ^ (11-2) = 11! / [2! * (11-2)] * (1 / 2) ^ 2 * (1 / 2) ^ 9 = 11 * 10 * 9 * 8 * 7 * 2 * 3 *…* 1 / [(2 * a) * (9 * 8 * 7 *…* 2 * 3 * a) * 0.5 * 55 ^ 11 = 1 / 2048 = 55/2048 Similarly, P (1 woman 11 participants) = 11C1 * (02/01) ^ 1 * (1 / 2) ^ (11-1) and P = 11 2048 (0 Women 11 participants) = 1 / 2048 for a total of N (no more than two women over 11 participants) = 55/2048 + 11/2048 + 1 / 2048 = 67/2048 = 0.0327

Monty Hall Problem: Numb3rs and 21