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Let S be the set of permutations of (1,2,3,4,5,6) 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?

 Dec 5, 2021
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The first term can be either 2, 3, or 5.

 

If the first term is 2, there are 5! = 5 x 4 x 3 x 2 x 1 = 120 ways to permutate the other terms.

Similarly, there are 120 ways if the first term is a 3 and there are 120 ways if the first term is a 5.

 

Therefore there is a total of 360 possible permutations for the denominator of the probability.

 

If the third term is a 2 and the first term is a 3, there are 4 x 3 x 2 x 1 = 24 ways.

Similarly, if the third term is a 2 and the first term is a 5, there are 24 ways.

 

Adding, we get 48 ways for the numerator of the probability.

 

The probability will be  24 / 360.

 Dec 6, 2021

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