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Paul and Jesse each choose a number at random from the first six primes. What is the probability that the sum of the numbers they choose is divisible by 3?

 Apr 16, 2022

Best Answer 

 #2
avatar+9675 
+1

Here are the first 6 primes: {2,3,5,7,11,13}

 

We can take a look at all the possibilities by making a table. 

 

235711132457913153568101416578101216187910121418201113141618222413151618202426

 

Blue numbers and red numbers represent the number that Paul chose and the number that Jesse chose respectively.

Pink numbers are those divisible by 3.

 

By trial-and-error, out of all 36 possible cases, 13 cases has a sum divisible by 3.

 

Therefore, 

 

Prob(sum divisible by 3)=1336

 Apr 16, 2022
 #1
avatar+37165 
+1

Assuming they can pick the same number

 

36 choices

2 7

2 13

5 7

5 13

7 11

3 3              the red ones can be reversed      11 choices out of 36

 

OOps...I see from Max answer I missed   11 13      13 11     so the answer would be the same    13 /36          (Thanx, max!)

 Apr 16, 2022
edited by Guest  Apr 16, 2022
 #2
avatar+9675 
+1
Best Answer

Here are the first 6 primes: {2,3,5,7,11,13}

 

We can take a look at all the possibilities by making a table. 

 

235711132457913153568101416578101216187910121418201113141618222413151618202426

 

Blue numbers and red numbers represent the number that Paul chose and the number that Jesse chose respectively.

Pink numbers are those divisible by 3.

 

By trial-and-error, out of all 36 possible cases, 13 cases has a sum divisible by 3.

 

Therefore, 

 

Prob(sum divisible by 3)=1336

MaxWong Apr 16, 2022

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