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Bob plays a game where, for some number n, he chooses a random integer between 0 and n-1, inclusive. If Bob plays this game for each of the first five prime numbers, what is the probability that the sum of the numbers he gets is greater than 0?

 Oct 17, 2020
 #1
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There is no way to get negative numbers, so the only way to not get a sum that is greater than 0 is when the sum = 0

 

\(\dfrac{1}{2} * \dfrac{1}{3} * \dfrac{1}{5} * \dfrac{1}{7} * \dfrac{1}{11} = \dfrac{1}{2310}\)

 

This is the chance that Bob gets 0 for all five games, since the chance of picking 0 out of n numbers is \(\dfrac{1}{n}\)

His chances of getting a sum of greater than 0 is

 

\(1 - \dfrac{1}{2310} = \dfrac{2309}{2310}\)

 
 Oct 17, 2020
 #2
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Bob plays a game where, for some number n, he chooses a random integer between 0 and n-1, inclusive. If Bob plays this game for each of the first five prime numbers, what is the probability that the sum of the numbers he gets is greater than 0?

 

I really do not understand the question. I probably have not interpreted it as intended.

 

The first 5 prime numbers are 2,3,5,7,11

All these numbers are positive and the integer  is between 0 and n-1 

So the integer cannot be 0

So the sum of the numbers must be greater than 0

P(greater than 0)=1

 
Melody  Oct 17, 2020

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