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Three distinct integers are chosen at random from the first 20 positive integers. Compute the probability that: (a) their sum is even; (b) their product is even.

 Feb 11, 2020
 #1
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Here's my best attempt

 

We  have   C (20, 3)   = 1140  possible different sets of 3 integers

 

 

(a)  the sum will be even when   we have

 

even + even + even        or

even + odd +  odd

 

In the first case  we want to choose  any  3  of the 10  even numbers  =  C (10, 3)  = 120

In the second  case we want to  choose  any 1 of 10 evens  = C (10, 1)  and  any 2 of the 10 odds = C(10,2) = 45 

 

So  the probability  is

 

[  C (10,3)  +  C(10,1) * C(10,2) ]               120  + 10 * 45            570

____________________________  =      _____________  =   ______  =   .5

         C (20, 3)                                                1140                      1140

 

 

(b)  the only way we can have an odd product is when all three  integers are odd.....there  are C (10,3)  = 120 sets of these

 

So.....the probability that we have an even product =

 

1   -   120  / 1140  =

 

[ 1140  -120 ]  / 1140  =

 

1020 / 1140  = 

 

17  /19

 

 

 

cool cool cool

 Feb 11, 2020

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