Probability
Compound Events
Grade 12
Question:
<p>Four fair dice <i>D</i><sub>1</sub>, <i>D</i><sub>2</sub>, <i>D</i><sub>3</sub> and <i>D</i><sub>4</sub> each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that <i>D</i><sub>4</sub> shows a number appearing on one of <i>D</i><sub>1</sub>, <i>D</i><sub>2</sub> and <i>D</i><sub>3</sub> is</p>
<p>(a) \(\frac{91}{216}\)</p>
<p>(b) \(\frac{108}{216}\)</p>
<p>(c) \(\frac{125}{216}\)</p>
<p>(d) \(\frac{127}{216}\)</p>
Step-by-Step Solution
Key Concept: Use the complement: find the probability that Dā does NOT match any of the other three dice, then subtract from 1.
<p><strong>Explanation:</strong> The probability that <i>D</i><sub>4</sub> shows a number that appears on at least one of <i>D</i><sub>1</sub>, <i>D</i><sub>2</sub>, <i>D</i><sub>3</sub> can be found using complementary counting. The complement is that <i>D</i><sub>4</sub> shows a number that does NOT appear on any of <i>D</i><sub>1</sub>, <i>D</i><sub>2</sub>, <i>D</i><sub>3</sub>. For each outcome of <i>D</i><sub>4</sub> (say value $k$), the probability that none of <i>D</i><sub>1</sub>, <i>D</i><sub>2</sub>, <i>D</i><sub>3</sub> show $k$ is $\left(\frac{5}{6}\right)^3$. Summing over all 6 outcomes of <i>D</i><sub>4</sub>: $P(\text{no match}) = \frac{125}{216}$. Therefore $P(\text{at least one match}) = 1 - \frac{125}{216} = \frac{91}{216}$.</p>
Correct Answer: A