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. \(\dfrac{1}{6}\)
• B. \(\dfrac{5}{18}\)
• C. \(\dfrac{7}{18}\)
• D. \(\dfrac{5}{6}\)

Hint:

For each possible first roll value \(a\), count how many values from 1 to 6 are exact multiples of \(a\), sum these counts, and divide by 36.

Answer 1

The Correct Option is C

Step 1: Set up the sample space.
Both rolls are independent and each takes a value in \(\{1,2,3,4,5,6\}\) with equal probability \(\frac{1}{6}\) per face. Let \(a\) be the first roll and \(b\) be the second roll. The total number of equally likely ordered pairs \((a,b)\) is \(6 \times 6 = 36\).

Step 2: Translate the event into a divisibility condition.
\(b\) is an integer multiple of \(a\) means \(a\) divides \(b\) exactly, i.e. \(b \bmod a = 0\), where \(b=a\) itself counts (since \(b=1 \times a\) is a valid integer multiple with multiplier 1).

Step 3: Count the favourable values of \(b\) for each value of \(a\).
\(a=1\): every \(b\) from 1 to 6 is a multiple of 1, giving 6 favourable values.
\(a=2\): multiples of 2 in range are 2, 4, 6, giving 3 favourable values.
\(a=3\): multiples of 3 in range are 3, 6, giving 2 favourable values.
\(a=4\): the only multiple of 4 in range is 4 itself (8 is out of range), giving 1 favourable value.
\(a=5\): the only multiple of 5 in range is 5 itself, giving 1 favourable value.
\(a=6\): the only multiple of 6 in range is 6 itself, giving 1 favourable value.

Step 4: Add up the favourable outcomes.
Total favourable pairs \(= 6+3+2+1+1+1 = 14\).

Step 5: Compute the probability.
\[ P = \frac{14}{36} = \frac{7}{18} \]

Step 6: Cross check against the options.
\(\frac{7}{18}\) matches option (C) exactly. Option (A) \(\frac{1}{6}=\frac{6}{36}\) undercounts by only including the \(a=1\) row and ignoring the equal case contributions from other rows. Option (B) \(\frac{5}{18}=\frac{10}{36}\) and option (D) \(\frac{5}{6}=\frac{30}{36}\) do not correspond to any consistent counting rule for this event and are ruled out by the direct enumeration above.
\[ \boxed{\dfrac{7}{18}} \]