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I play a game where, for any random number x, I choose a random integer between 0 and x-1, inclusive. If I play this game for each of the first four prime numbers, what is the probability that the sum of the numbers I get is greater than 0?

 

Sorry, the first time I posted this part of it didn't post...

 Jul 13, 2018
edited by MathCuber  Jul 13, 2018

Best Answer 

 #1
avatar+33603 
+2

Tackle this by first calculating the probability, p, that you get 0 for each of the four cases. The required probability is then simply 1 - p.

 

The first four prime numbers are 2, 3, 5, 7.

 

The corresponding probabilities of choosing 0 are: 1/2, 1/3, 1/5, 1/7.

 

Multiply these together to find p. Subtract the result from 1 to get the probability that the sum of the chosen numbers is greater than 0.

 

I’ve assumed that a “game” consists of selecting each of the first four primes and choosing a random integer between 0 and one less than that prime.  It’s possible you mean the primes to be chosen at random also (from the first four) - it isn’t clear to me which scenario you want!

 Jul 13, 2018
 #1
avatar+33603 
+2
Best Answer

Tackle this by first calculating the probability, p, that you get 0 for each of the four cases. The required probability is then simply 1 - p.

 

The first four prime numbers are 2, 3, 5, 7.

 

The corresponding probabilities of choosing 0 are: 1/2, 1/3, 1/5, 1/7.

 

Multiply these together to find p. Subtract the result from 1 to get the probability that the sum of the chosen numbers is greater than 0.

 

I’ve assumed that a “game” consists of selecting each of the first four primes and choosing a random integer between 0 and one less than that prime.  It’s possible you mean the primes to be chosen at random also (from the first four) - it isn’t clear to me which scenario you want!

Alan Jul 13, 2018
 #2
avatar+537 
+2

You assumed right, Thank you!

MathCuber  Jul 14, 2018

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