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 number appearing in the second roll is an
integer multiple of the number appearing in the first roll is __________

• A. 1 6
• B. 5 18
• C. 7 18
• D. 5 6

Hint:

For each value of the first roll \(a\), count how many of the six second-roll faces are multiples of \(a\); sum these counts and divide by 36 to get \(7/18\).

Answer 1

The Correct Option is C

Instead of listing all 36 outcomes, organize the count by fixing the first roll and asking "how many second-roll values make it a multiple?"

Since both rolls come from an unbiased die with faces $\{1,2,3,4,5,6\}$, the sample space has $6 \times 6 = 36$ equally likely ordered pairs $(a,b)$.

Go through each value of the first roll $a$ and count how many values of $b$ (from 1 to 6) are multiples of $a$:
- $a=1$: every face 1-6 is a multiple of 1, giving 6 matches.
- $a=2$: only 2, 4, 6 are multiples of 2 within the range, giving 3 matches.
- $a=3$: only 3 and 6 qualify, giving 2 matches.
- $a=4$: only 4 itself qualifies (8 is out of range), giving 1 match.
- $a=5$: only 5 itself qualifies, giving 1 match.
- $a=6$: only 6 itself qualifies, giving 1 match.

Adding these up: $6+3+2+1+1+1 = 14$ favorable ordered pairs out of 36 total.

Simplify: $\dfrac{14}{36} = \dfrac{7}{18}$.

$\text{Probability} = \dfrac{7}{18}$

Correct option: (C)