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. \(\dfrac{3}{36}\)
• B. \(\dfrac{13}{36}\)
• C. \(\dfrac{15}{36}\)
• D. \(\dfrac{19}{36}\)

Hint:

List the possible sums from 2 to 12, keep only the prime ones, and count the ordered pairs that give each of those sums.

Answer 1

The Correct Option is C

Step 1: Count outcomes, this time organised by the first roll.
There are $6\times6=36$ equally likely pairs $(a,b)$ for the two rolls. Instead of grouping by the sum, go through each value of the first roll $a$ and count how many values of $b$ make $a+b$ prime.

Step 2: First roll is 1.
$a=1$: $b$ must make $1+b$ prime. Checking $b=1,\dots,6$ gives sums $2,3,4,5,6,7$. The primes among these are $2,3,5,7$, from $b=1,2,4,6$. That is 4 values.

Step 3: First roll is 2.
$a=2$: sums are $3,4,5,6,7,8$. Primes are $3,5,7$, from $b=1,3,5$. That is 3 values.

Step 4: First roll is 3.
$a=3$: sums are $4,5,6,7,8,9$. Primes are $5,7$, from $b=2,4$. That is 2 values.

Step 5: First roll is 4.
$a=4$: sums are $5,6,7,8,9,10$. Primes are $5,7$, from $b=1,3$. That is 2 values.

Step 6: First roll is 5.
$a=5$: sums are $6,7,8,9,10,11$. Primes are $7,11$, from $b=2,6$. That is 2 values.

Step 7: First roll is 6.
$a=6$: sums are $7,8,9,10,11,12$. Primes are $7,11$, from $b=1,5$. That is 2 values.

Step 8: Add up all six counts.
\[ 4+3+2+2+2+2=15 \]
This matches the count found by grouping by sum, confirming 15 favourable outcomes out of 36.
\[ \boxed{\dfrac{15}{36}} \]