Let S be the set of permutations of (1,2,3,4,5) whose first term is a prime.
If we choose a permutation at random from S, what is the probability that the third term is equal to 2?
thx
+23 The set S contains 72 numbers. Since the restriction is that the first term must be a prime, only 2, 3, and 5 work. 3 * 4 * 3 * 2 * 1 = 72
Now, if we need the third term to equal 2, we can get 2 * 3 * 1 * 2 * 1, which equals 12.
12/72 = 1/6
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+23 The set S contains 72 numbers. Since the restriction is that the first term must be a prime, only 2, 3, and 5 work. 3 * 4 * 3 * 2 * 1 = 72
Now, if we need the third term to equal 2, we can get 2 * 3 * 1 * 2 * 1, which equals 12.
12/72 = 1/6
:>
