+0  
 
0
574
2
avatar

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.

Guest Apr 5, 2015

Best Answer 

 #1
avatar+81077 
+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
Sort: 

2+0 Answers

 #1
avatar+81077 
+5
Best Answer

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
avatar
0

I'm just adding to CPhill's answer

14/35=2/5

Guest Aug 19, 2016

30 Online Users

avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details