# Choose the Dress

A mother bought three dress for her triplets daughters(one for each) and put the dresses in the dark. One by one the girls come and pick a dress.

What is the probability that no girl will choose her own dress?

Let’s assume that the three dresses are indistinguishable and that each girl picks a dress at random. There are 3! = 6 ways that the girls can pick the dresses, since each girl can pick from any of the three dresses, and there are 3 ways to assign the first girl’s choice, 2 ways to assign the second girl’s choice (since one dress is already chosen), and 1 way to assign the third girl’s choice (since two dresses are already chosen).

Out of these 6 possible ways, we need to count the number of ways that no girl picks her own dress. We can use the principle of inclusion-exclusion to do this.

Let A, B, and C denote the events that the first, second, and third girl, respectively, picks her own dress. Then, we want to calculate the probability of the event not(A or B or C), which is the complement of the event A or B or C. By the principle of inclusion-exclusion, we have:

P(not(A or B or C)) = 1 – P(A or B or C) = 1 – (P(A) + P(B) + P(C) – P(A and B) – P(A and C) – P(B and C) + P(A and B and C))

Since each girl picks a dress at random, we have:

P(A) = 1/3 P(B) = 1/2 (since one dress is already chosen, the second girl has only 2 choices) P(C) = 1 (since there is only one dress left for the third girl)

Also, we have:

P(A and B) = 1/3 * 1/2 = 1/6 P(A and C) = 1/3 * 1/1 = 1/3 P(B and C) = 1/2 * 1/1 = 1/2

Finally, we have:

P(A and B and C) = 1/3 * 1/2 * 1/1 = 1/6

Substituting these values into the formula above, we get:

P(not(A or B or C)) = 1 – (1/3 + 1/2 + 1 – 1/6 – 1/3 – 1/2 + 1/6) = 1/3

Therefore, the probability that no girl will choose her own dress is 1/3.

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