GATE CSE 2026 | Set 2 | GA | Question: 10

Question

An unbiased six-faced dice whose faces are marked with numbers $1,2,3,4,5$, and $6$ is rolled twice in succession and the number on the top face is recorded each time. The probability that the sum of the two recorded numbers is a prime number is $\_\_\_\_\_\_\_$

• A. $\frac{3}{36}$
• B. $\frac{13}{36}$
• C. $\frac{15}{36}$
• D. $\frac{19}{36}$

Comments

face sum is prime

possible sum is 2,3,5,7,11

2 = (1,1)

3 = (2,1),(1,2)

5 = (1,4),(4,1),(2,3),(3,2)

7 = (1,6),(6,1),(2,5),(5,2),(3,4),(4,3)

11 = (5,6),(6,5)

total case  = 1+2+4+6+2 = 15

answer = 15 / 36

Answer 1

Total outcomes from rolling two die in succession is 36.

The sum of two top faces in this event will range from [2-12] as (1,1) = 2, and largest possible is (6,6) = 12.

In these outcomes we can just count the number of prime numbers output as it is a small sample space.

The prime numbers between 2-12 are 2,3,5,7,11.

2 : (1,1)

3 : (1,2) (2,1)

5 : (1,4) (2,3) (3,2) (4,1)

7 : (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

11 : (5,6) (6,5)

Counting them leads to 1+2+4+6+2 = 15.

Therefore, probabilty of having a sum being a prime number is = Favourable outcomes / total outcomes.

= 15/36.

 

Option C

Comments

Used exact same steps

Answer 2

When a six-faced die is rolled twice, the total number of possible outcomes in the sample space $S$ is:

$n(S) = 6 \times 6 = 36$

here we use six-sided dice as a pair, then the minimum possible sum will be $1+1=2$, and the maximum will be $6+6=12$. so the required prime numbers between 2 and 12 are: $\{2, 3, 5, 7, 11\}$.

Now, we have to find the pairs $(x, y)$ that result in these sums:

• Sum = 2: $(1, 1) \rightarrow \mathbf{1 \text{ possibilities}}$

• Sum = 3: $(1, 2), (2, 1) \rightarrow \mathbf{2 \text{ possibilities}}$

• Sum = 5: $(1, 4), (4, 1), (2, 3), (3, 2) \rightarrow \mathbf{4 \text{ possibilities}}$

• Sum = 7: $(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) \rightarrow \mathbf{6 \text{ possibilities}}$

• Sum = 11: $(5, 6), (6, 5) \rightarrow \mathbf{2 \text{ possibilities}}$

therefore total favorable outcome is: $n(E) = 1 + 2 + 4 + 6 + 2 = 15$

$P(\text{Sum is Prime}) = \frac{n(E)}{n(S)} = \frac{15}{36}$

Option $(C)$ is correct.