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:

Fix the first roll and count how many second-roll values are multiples of it, then add up over all six first-roll values.

Answer 1

The Correct Option is C

Step 1: Set up the sample space.
The dice is rolled twice, and each roll is independent with six equally likely outcomes. So the total number of ordered pairs \((x,y)\), where \(x\) is the first roll and \(y\) is the second roll, is \[ 6\times6=36 \]

Step 2: State the event we need.
We want the second roll \(y\) to be an integer multiple of the first roll \(x\), meaning \(y=kx\) for some positive whole number \(k\). Since both \(x\) and \(y\) lie between \(1\) and \(6\), we check each value of \(x\) separately and count how many values of \(y\) work.

Step 3: Count the favorable \(y\) values for each \(x\).
For \(x=1\): every value \(y=1,2,3,4,5,6\) is a multiple of \(1\), giving \(6\) values. For \(x=2\): the multiples of \(2\) up to \(6\) are \(2,4,6\), giving \(3\) values. For \(x=3\): the multiples of \(3\) up to \(6\) are \(3,6\), giving \(2\) values. For \(x=4\): the only multiple of \(4\) up to \(6\) is \(4\) itself, giving \(1\) value. For \(x=5\): the only multiple of \(5\) up to \(6\) is \(5\) itself, giving \(1\) value. For \(x=6\): the only multiple of \(6\) up to \(6\) is \(6\) itself, giving \(1\) value.

Step 4: Add up the favorable outcomes.
\[ 6+3+2+1+1+1=14 \]

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

Step 6: Rule out the other options.
\(\dfrac{1}{6}=\dfrac{6}{36}\) is too small, since it undercounts many valid pairs like \((2,4)\) or \((3,6)\). \(\dfrac{5}{18}=\dfrac{10}{36}\) misses a few valid pairs. \(\dfrac{5}{6}=\dfrac{30}{36}\) is far too big, since not every pair \((x,y)\) has \(y\) a multiple of \(x\); for instance \((4,5)\) does not qualify.

Final Answer:
The probability is \(\dfrac{7}{18}\). \[ \boxed{\dfrac{7}{18}} \]