Question

A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first?

• A. $\tfrac{5}{36}$
• B. $\tfrac{8}{36}$
• C. $\tfrac{15}{36}$
• D. $\tfrac{21}{36}$
• E. None of the above

Hint:

For two dice comparisons, the cases $b>a$ and $b<a$ are symmetric. Hence each has $\frac{15}{36}$ probability, while $b=a$ has $\frac{6}{36}$. This symmetry often speeds up such problems.

Answer 1

The Correct Option is C

Step 1: Total outcomes.
When a fair die is rolled twice, each roll has 6 outcomes. Thus total outcomes: \[ 6 \times 6 = 36 \]

Step 2: Favorable outcomes (second $>$ first).
We count ordered pairs \((a,b)\) with \(b>a\) where \(a\) = first roll, \(b\) = second roll. - If \(a=1\): \(b=2,3,4,5,6 \;\Rightarrow 5\) possibilities. - If \(a=2\): \(b=3,4,5,6 \;\Rightarrow 4\) possibilities. - If \(a=3\): \(b=4,5,6 \;\Rightarrow 3\) possibilities. - If \(a=4\): \(b=5,6 \;\Rightarrow 2\) possibilities. - If \(a=5\): \(b=6 \;\Rightarrow 1\) possibility. - If \(a=6\): \(b>a\) is impossible \(\;\Rightarrow 0\). Total favorable outcomes: \[ 5+4+3+2+1=15 \] 

Step 3: Probability.
\[ P=\frac{\text{favorable outcomes}}{\text{total outcomes}} =\frac{15}{36} \] 

Final Answer:
\[ \boxed{\text{(C) } \tfrac{15}{36}} \]