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Paul and Jesse each choose a number at random from the first six primes. What is the probability that the sum of the numbers they choose is even?

TheMathCoder  Apr 26, 2018
 #1
avatar+87333 
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First 6 primes  =

 

2, 3, 5, 7, 11, 13

 

Notice that each can choose any one of six numbers....so...the total number of possibilities is  6 * 6  = 36

 

But  the only odd sums occur when one of the chooses a 2 and the other chooses an odd prime....

 

So for   (Paul, Jessie)....these will occur  as  ( 2, odd)  or ( odd, 2) 

 

And there are 10 total sets where an odd sum occurs...so....the probability of an even sum is :

 

26/36  =    13/18

 

 

 

cool cool cool

CPhill  Apr 26, 2018
edited by CPhill  Apr 26, 2018
edited by CPhill  Apr 26, 2018
edited by CPhill  Apr 26, 2018
 #2
avatar+290 
+4

Thanks CPhill!

TheMathCoder  Apr 26, 2018

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