Read e-book online An introduction to probability theory and its applications PDF By William Feller

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12). Thus, we have seen that it is possible to think of the black and white marbles as being indistinguishable within themselves in this problem, and the same value for P(exactly 2 black) is obtained; however, the computation is no longer carried out over an original basic sample space of equally likely outcomes. For example, on the first draw one would just have the two possible outcomes, B and W, although these two outcomes obviously would not be equally likely, but rather P(B) = 10/30 and P(W) = 20/30.

It would be necessary to verify P(A n B) = P(A)P(B), P(A n B n C) P(B n C) = P(B)P(C), and also P(A n C) = P(A)P(C), = P(A)P(B)P(C). The following examples show that pairwise independence does not imply this last threeway factorization and vice versa. 7 A box contains eight tickets, each labeled with a binary number. Two are labeled 111, two are labeled 100, two 010, and two 001. An experiment consists of drawing one ticket at random from the box. 9. It follows that P(A) = P(B) = P(C) 4/8 = 1/2 and that P(A n B) = P(A n C) P(B n C) = 2/8 = 1/4, thusAB, and C arçpa1r wise independent.

12 as before. 11. 6 using combinations. , n taken three at a time. Equivalently, this would be the collection of all subsets of size 3 from the set {1. ,, n}, of which there are (n'\ n(n lXn-2) n! , {n - 2, n - 1, n}. As usual, no distinction should be made of subsets that list the elements in a different order. The resulting probability is (n-2) 6 [n(n - 1)(n - 2)16] as before. This shows that some problems can be solved using either combinations or permutations. Usually, if there is a choice, the combination approach is simpler because the sample space is smaller.