Probability
Combinatorial Probability
Grade 12
Question:
<p>If three integers are selected from the set of the first 20 natural numbers, the probability that their product is a multiple of 3, is</p>
<p>(a) \(\frac{3}{28}\)</p>
<p>(b) \(\frac{2}{28}\)</p>
<p>(c) \(\frac{7}{28}\)</p>
<p>(d) \(\frac{194}{285}\)</p>
Step-by-Step Solution
Key Concept: Use complementary counting: product is multiple of 3 if at least one factor is divisible by 3; count selections with at least one such number.
<p><strong>Solution:</strong> Total number of ways of selecting 3 integers from 20 natural numbers: $n(S) = \binom{20}{3} = 1140$</p><p>Their product is a multiple of 3 means at least one number is divisible by 3. The numbers divisible by 3 among the first 20 natural numbers are 3, 6, 9, 12, 15, and 18 (6 numbers).</p><p>Number of ways of selecting at least one multiple of 3: $n(E) = \binom{6}{1} \cdot \binom{14}{2} + \binom{6}{2} \cdot \binom{14}{1} + \binom{6}{3} = 776$</p><p>Required probability: $P(E) = \frac{n(E)}{n(S)} = \frac{776}{1140} = \frac{194}{285}$</p>
Correct Answer: d