Five numbers taken out from numbers 1-30 and arrange them in ascending order. The probability that the third number will be 20 is

\(\frac{^{19}C_2×^{10}C_2}{^{30}C_5}\)

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 = $\frac{{^{19}C}_2 × {^{10}C}_2}{{^{30}C}_5}⇒\frac{19×18}{2}×\frac{10×9}{2}×\frac{5×4×3×2×1}{30×29×28×27×26}=\frac{285}{5278}$