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A cube with its sides numbered 1 through 6 is rolled twice, first land
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09 Jul 2021, 08:37
09 Jul 2021, 08:37
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A cube with its sides numbered 1 through 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 through 6, what is the probability that a + b is prime?
(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9
(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9
A cube with its sides numbered 1 through 6 is rolled twice, first land
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10 Jul 2021, 03:50
10 Jul 2021, 03:50
1
Solution:
The total number of possibilities will be 6 x 6=36.
And out of this 36 number only 2(1+1) and 12(6+6) can be computed in 1 ways rest all others can be computed in different ways.
Thus, the range of the numbers also lie between 2-12.
Prime numbers between 2-12= 2, 3, 5, 7, 11
2 can be formed in 1 way= 1+1=2
3 can be formed in 2 ways= 1+2 & 2+1
5 can be formed in 4 ways= 1+4, 4+1, 2+3, 3+2
7 can be formed in 6 ways= 1+6, 6+1, 2+5, 5+2, 3+4, & 4+3
11 can be formed in 2 ways= 6+5 & 5+6
Therefore the above sums up to 15 possible ways
Probability that a + b is prime= \(\frac{Total number of ways the sum is prime}{ Total number of all possibilities}\)=
=\(\frac{15}{36}\)=\(\frac{5}{12}\)
IMO C
Hope this helps!
The total number of possibilities will be 6 x 6=36.
And out of this 36 number only 2(1+1) and 12(6+6) can be computed in 1 ways rest all others can be computed in different ways.
Thus, the range of the numbers also lie between 2-12.
Prime numbers between 2-12= 2, 3, 5, 7, 11
2 can be formed in 1 way= 1+1=2
3 can be formed in 2 ways= 1+2 & 2+1
5 can be formed in 4 ways= 1+4, 4+1, 2+3, 3+2
7 can be formed in 6 ways= 1+6, 6+1, 2+5, 5+2, 3+4, & 4+3
11 can be formed in 2 ways= 6+5 & 5+6
Therefore the above sums up to 15 possible ways
Probability that a + b is prime= \(\frac{Total number of ways the sum is prime}{ Total number of all possibilities}\)=
=\(\frac{15}{36}\)=\(\frac{5}{12}\)
IMO C
Hope this helps!
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Re: A cube with its sides numbered 1 through 6 is rolled twice, first land
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15 Oct 2022, 22:12
15 Oct 2022, 22:12
Given that A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. and We need to find If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?
As we are rolling the cube twice => Number of cases = \(6^2\) = 36
First roll gives a and second roll gives b and we need to find the Probability that a+b is prime => Sum of the two rolls is prime
We know that out of the numbers from 2(1+1) to 12 (6+6) we have only five prime numbers and they are 2, 3, 5, 7 and 11
(Watch this video if you are not aware of Prime Numbers)
Following are the possible cases in which sum of the two rolls is a prime number
(1,1), (1,2), (1,4), (1,6)
(2,1), (2,3), (2,5)
(3,2), (3,4)
(4,1), (4,3)
(5,2), (5,6)
(6,1), (6,5)
=> 15 possibilities
=> Probability that a + b is prime = \(\frac{15}{36}\) = \(\frac{5}{12}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
As we are rolling the cube twice => Number of cases = \(6^2\) = 36
First roll gives a and second roll gives b and we need to find the Probability that a+b is prime => Sum of the two rolls is prime
We know that out of the numbers from 2(1+1) to 12 (6+6) we have only five prime numbers and they are 2, 3, 5, 7 and 11
(Watch this video if you are not aware of Prime Numbers)
Following are the possible cases in which sum of the two rolls is a prime number
(1,1), (1,2), (1,4), (1,6)
(2,1), (2,3), (2,5)
(3,2), (3,4)
(4,1), (4,3)
(5,2), (5,6)
(6,1), (6,5)
=> 15 possibilities
=> Probability that a + b is prime = \(\frac{15}{36}\) = \(\frac{5}{12}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
