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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

 May 28, 2021

Best Answer 

 #1
avatar+23 
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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

 

:>

 May 28, 2021
 #1
avatar+23 
+2
Best Answer

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

 

:>

ExpireNight May 28, 2021

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