Latest Infosys Aptitude Question SOLUTION: Two dice are thrown simultaneously. What is the probability that the sum of the numbers shown on the two dices will be a prime number?

• Placement Papers
• /
• Infosys
• /
• Numerical Ability
• /
• Probability

Infosys Question Papers: Infosys Interview Experience(10) INFOSYS Company Profile

(#M40153919) INFOSYS Numerical Ability question Keep an EYE

Two dice are thrown simultaneously. What is the probability that the sum of the numbers shown on the two dices will be a prime number?

Asked In Infosys kanmani (8 years ago)

Unsolved Read Solution (15)

Is this Puzzle helpful?   (25)   (10) Submit Your Solution Probability

Read Solution (15) :

animesh kumar  (8 years ago)

number of sample solution=36
number of possible outcomes=(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
p=15/36

Is this solution Helpfull? Yes (25) | No (2)  

pavithran  (8 years ago)

number of outcomes = 15 (sum of prime numbers)
possible outcomes = 36
p(A) = 15/36 = 5/12

Is this solution Helpfull? Yes (6) | No   

sai prathyusha  (7 years ago)

two dice thrown means 6*6=36
sum of two dice is prime number means 2,3,5,7,11......,
no of events={(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
probability=no of events/36=15/36=5/12

Is this solution Helpfull? Yes (3) | No   

Rajaboina Radhika  (6 years ago)

sum is 2,3,5,7 and 11
possible combinations are
1,1
1,2
2,1
1,4,
2,3,
3,2,
4,1
,5,6,
6,5,
1,6,
2,5,
3,4,
6,1,
5,2,
4,3
hence the probability to get sum as prime on the two dice is 15/36

Is this solution Helpfull? Yes (3) | No (1)  

meghana  (8 years ago)

total number of solutions =36
no of possible outcomes = (1,1),(1,2),(1,4),(1,6),(2,3),(2,5),(3,4),(5,6) = 8
remaining (2,1),(4,1),(6,1),(3,2),(5,2),(4,3),(6,5) are repeated
probability = 8/36

Is this solution Helpfull? Yes (2) | No (13)  

Sourabh Sharma  (8 years ago)

7/18.As there r 36 total cases..And 14 favourable cases .

Is this solution Helpfull? Yes (2) | No (4)  

mashhud ansari  (8 years ago)

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E = { (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) }
n(E) = 15.

P(E) = n(E) = 15 = 5 .
n(S) 36 12

Is this solution Helpfull? Yes (1) | No   

madesh  (8 years ago)

sum of dice= 2 3 4 5 6 7 8 9 10 11 12
no.of possible= 1 2 3 4 5 6 5 4 3 2 1

Is this solution Helpfull? Yes (1) | No   

Seepala manju bhargavi  (5 years ago)

when u throw two dice .there will be 15 prime numbers ,so 36/15 = 5/12

Is this solution Helpfull? Yes (1) | No   

pavankumar  (8 years ago)

The Possible Combinations are (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 and total outcome 6*6=36, So answer will be 15/36=5/12.

Is this solution Helpfull? Yes | No (1)  

Dhvani Agola  (8 years ago)

total number of outcomes=6*6=36
there are pairs which have prime sum=(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)
total number of above pairs=15
so,p=15/36 = 5/12

Is this solution Helpfull? Yes | No   

Syed Farhaz Ali  (8 years ago)

Select 6 men from 12 i.e 12C6 =926
Now select 1 women from 3 i.e 3C1 =3
then add 926 + 3 = 929,which is the answer

Is this solution Helpfull? Yes | No (2)  

ayushi  (8 years ago)

there is probability of getting sum prime case 1=1+2,1+4
case 2= 2+3,2+1
case 3 =3+2
case 4=4+1
probability=6/12=1/2

Is this solution Helpfull? Yes | No (2)  

rajan kumar mehta  (7 years ago)

sum i.e prime=2 3 5 7 11
total combination =2,3,4.................12
required probablity=5/11

Is this solution Helpfull? Yes | No (2)  

ragu  (5 years ago)

Rolling 2 dice gives a total of 36 possible outcomes.
Here is the sample space:


Prob of getting a sum of atleast 6 = Prob of getting sum >=6

Following table shows the possible cases(enclosed in orange box) for getting the required sum.


Prob[Required Sum] =
P[S=6]+P[S=7]+P[S=8]+P[S=9]+P[S=10]+P[S=11]+P[S=12]

= 5/36+6/36+5/36+4/36+3/36+2/36+1/36

= 26/36

= 0.722

Therefore, Required probability = 0.722

Is this solution Helpfull? Yes | No (3)  

Please Login to Read Solution.