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. \(\frac{19}{36}\)

Hint:

List all sums from 2 to 12 that are prime (2, 3, 5, 7, 11), count the dice-pair combinations for each, and divide the total by 36.

Answer 1

The Correct Option is C

Instead of listing prime sums directly, build the full sample space mentally and mark the prime totals.

Setup: Rolling two dice twice in succession gives ordered pairs \((a,b)\) with \(a, b \in \{1,2,3,4,5,6\}\), so there are \(36\) equally likely outcomes in total.

Identify primes in range: the sum \(a+b\) can be anywhere from \(2\) to \(12\). Checking each value for primality: \(2,3,5,7,11\) are prime, while \(4,6,8,9,10,12\) are composite.

Count outcomes per prime sum using the standard triangular counting rule (number of ways to get sum \(s\) with two dice is \(s-1\) for \(s \le 7\) and \(13-s\) for \(s \ge 7\)): \(s=2 \Rightarrow 1\) way, \(s=3 \Rightarrow 2\) ways, \(s=5 \Rightarrow 4\) ways, \(s=7 \Rightarrow 6\) ways, \(s=11 \Rightarrow 2\) ways.

Adding these: \(1+2+4+6+2 = 15\) favourable outcomes out of \(36\).

So the required probability is \(\dfrac{15}{36}\), which is option C.