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A cube with its sides numbered 1 6 is rolled twice, first landing on

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A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink] New post   27 Oct 2018, 10:29
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A cube with its sides numbered 1— 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 — 6, what is the probability that a + b is prime?

A) 0

B) \(\frac{1}{12}\)

C) \(\frac{5}{12}\)

D) \(\frac{7}{18}\)

E) \(\frac{4}{9}\)

Weekly Quant Quiz #6 Question No 10



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pandeyashwin
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Re: A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink] New post   27 Oct 2018, 13:21
Prime nos between 1-12 : 2 , 3 , 5 , 7, 11
for 2 : (1,1) : 1/6*1/6 = 1/36
3 : (2,1), (1,2): 2 * 1/6 * 1/6 = 1/18
5 : (1,4) and (2,3) = 4*1/6*1/6 = 2/18
7 : (1,6) , (2,5) , (3,4) = 6*1/6*1/6 = 3/18
11 : (6,5) = 2*1/6*1/6 = 1/18

Total = 15/36 = 5/12
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Re: A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink] New post   28 Oct 2018, 01:12
total outcome 36
prime comb 15
2,3,5,7
3,5,7
5,7
5,7
7,11
7,11

hence -> 15/36 --> 5/12 (C)
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Re: A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink] New post   28 Oct 2018, 02:54
Is it C?
This is how i approached the question. As I am not sure at all please let me know if there's a mistake.

If a=1, then possible outcomes of b to make a + b prime number is 1, 2, 4, 6 => p= (1/6)*(4/6)
then a=2, a=3 and go on.

I got 15/36 = 5/12
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Re: A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink] New post   28 Oct 2018, 21:41
We can count starting from 1 adding every other number to it as long as it sums upto a prime number.
Like:-
1, 2, 3, 4, 5, 6
1+1=2
1+2=3
1+4=5
1+6=7
2+1=3
2+3=5
2+5=7
3+2=5
3+4=7
4+1=5
4+3=7.......so on
Total favorable outcomes are 15
and total outcomes are 36
So probability of two numbers adding upto a prime number is 15/36=5/12.

Hope I am right !
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Re: A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink] New post   02 Oct 2019, 08:13
Quote:
A cube with its sides numbered 1— 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 — 6, what is the probability that a + b is prime?

A) 0

B) \(\frac{1}{12}\)

C) \(\frac{5}{12}\)

D) \(\frac{7}{18}\)

E) \(\frac{4}{9}\)



1 - 1/2/3/4/5/6
2,3,5,7

2 - 1/2/3/4/5/6
3,5,7

3 - 1/2/3/4/5/6
5,7

4 - 1/2/3/4/5/6
5,7

5 - 1/2/3/4/5/6
7,11

6 - 1/2/3/4/5/6
7,11

15/36
=5/12

Note: Since a and b are different in sequence so 7 in case 6 is different from 7 in case 5.

Thank you!

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Re: A cube with its sides numbered 1 6 is rolled twice, first landing on [#permalink]
02 Oct 2019, 08:13
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