Five numbers taken out from numbers 1-30 and arrange them in ascending order. The probability that the third number will be 20 is
Answer & explanation
Correct answer: option 2
Total no. of ways in which 5 tickets can be drawn = $n(5) = {^{30}C}_5$
The tickets are arranged in the form T1, T2, T3 (= 20), T4, T5
Where T1, T2 ∈ {1, 2, 3, ..., 19} and T4, T5 ∈ {21, 22, ..., 30}
∴ No. of favourable cases =${^{19}C}_2 ×1× {^{10}C}_2$
∴ Required probability = $\frac{{^{19}C}_2 × {^{10}C}_2}{{^{30}C}_5}⇒\frac{19×18}{2}×\frac{10×9}{2}×\frac{5×4×3×2×1}{30×29×28×27×26}=\frac{285}{5278}$