Probability
Combinatorial Selection
Grade 12
Question:
<p>Choosing <i>{x, y, z} ⊂ S</i>, such that <i>x, y, z</i> are not consecutive, is</p>
<p>(a) <i>17/70</i></p>
<p>(b) <i>34/70</i></p>
<p>(c) <i>51/70</i></p>
<p>(d) <i>34/35</i></p>
Step-by-Step Solution
Key Concept: Use complementary counting: subtract the number of ways to choose three consecutive numbers from the total.
<p><strong>Solution:</strong> Given, <i>x, y, z</i> are not consecutive.</p><p>Number of favourable ways = $21 - \binom{19}{3}$</p><p>= $\binom{21}{3} - \binom{19}{3} = 1330 - \frac{19 \times 18 \times 17}{1 \times 2 \times 3}$</p><p>= $1330 - 969 = 361$</p><p>Wait, recalculating: Number of favourable ways = $\binom{19}{3} = \frac{19 \times 18 \times 17}{6} = 969$</p><p>Total number of ways = $\binom{21}{3} = 1330$</p><p>Required probability = $\frac{969}{1330} = \frac{51}{70}$</p>
Correct Answer: C