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# Four prime numbers are randomly selected without replacement from the first ten prime numbers. What is the probability that the sum of the f

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Four prime numbers are randomly selected without replacement from the first ten prime numbers. What is the probability that the sum of the four selected numbers is odd? Express your answer as a common fraction.

Apr 5, 2015

#1
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The number of total possibilities is C(10,4 ) = 210

But the sum of any four of the first ten prime numbers is odd only if 2 is included.

So, this means that, out of the other nine, we want to choose any three of them.

So C(9,3)  = 84

So...the probability that the sum of any four of the first ten prime numbers is odd is given by ;

84 / 210 =  28/70 = 14/35   Apr 6, 2015

#1
+5

The number of total possibilities is C(10,4 ) = 210

But the sum of any four of the first ten prime numbers is odd only if 2 is included.

So, this means that, out of the other nine, we want to choose any three of them.

So C(9,3)  = 84

So...the probability that the sum of any four of the first ten prime numbers is odd is given by ;

84 / 210 =  28/70 = 14/35   CPhill Apr 6, 2015
#2
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