Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

1³/₂₀
2²⁹/₃₄
3⁴⁷/₁₀₀
4¹³/₁₀₂

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let S be the sample space . \\ Then, n (S) =^{52} C_{2} = \frac{(52 \times 51)}{(2 \times 1)} = 1326. \\ Let E = event of getting 1 spade and 1 heart . \\ ∴ n (E) = number of ways of choosing 1 spade \\ out of 13 and 1 heart out of 13 \\ = (^{13} C_{1} \times^{13} C_{1}) \\ = (13 \times 13) \\ = 169. \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{169}{1326} = \frac{13}{102} . \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{S}}\,{\text{be}}\,{\text{the}}\,{\text{sample}}\,{\text{space}}{\text{.}} \cr & {\text{Then}},\,n\left( S \right) = {}^{52}{C_2} = \frac{{\left( {52 \times 51} \right)}}{{\left( {2 \times 1} \right)}} = 1326. \cr & {\text{Let}}\,{\text{E}}\,{\text{ = }}\,{\text{event}}\,{\text{of}}\,{\text{getting}}\,{\text{1}}\,{\text{spade}}\,{\text{and}}\,{\text{1}}\,{\text{heart}}{\text{.}} \cr & \therefore n\left( E \right) = {\text{number}}\,{\text{of}}\,{\text{ways}}\,{\text{of}}\,{\text{choosing}}\,{\text{1}}\,{\text{spade}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{out}}\,{\text{of}}\,{\text{13}}\,{\text{and}}\,{\text{1}}\,{\text{heart}}\,{\text{out}}\,{\text{of}}\,{\text{13}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {{}^{13}{C_1} \times {}^{13}{C_1}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {13 \times 13} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 169. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{169}}{{1326}} = \frac{{13}}{{102}}. \cr}$