Question

Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that either both are black or both are kings.

Answer 1

Let S be the sample space.
Then n(S) = number of ways of drawing two cards out of 52 = ⁵²C₂
Let E₁ = event in which both the cards are black cards
and E₂ = event in which both the cards are kings.
Then (E₁∩ E₂) = event of getting two black kings
i.e. n(E₁) = number of ways of drawing two black cards out of the black cards = ²⁶C₂
n(E₂) = number of ways of drawing two kings out of four kings = ⁴C₂
∴ n(E₁∩ E₂) = number of ways of drawing two black kings out of two kings = ²C₂ = 1
Thus,

\[P\left( E_1 \right) = \frac{n\left( E_1 \right)}{n\left( S \right)} = \frac{^{26}{}{C}_2}{^{52}{}{C}_2}; P\left( E_2 \right) = \frac{n\left( E_2 \right)}{n\left( S \right)} = \frac{^{4}{}{C}_2}{^{52}{}{C}_2}\] and  \[P\left( E_1 \cap E_2 \right) = \frac{n\left( E_1 \cap E_2 \right)}{n\left( S \right)} = \frac{1}{^{52}{}{C}_2}\]

∴ P(drawing both red cards or both kings) = P(E₁or  E₂)
                                                                = P(E₁∪ E₂)
                                                                = P(E₁) + P(E₂) -P( E₁∩ E₂)
                                                                =\[\left( \frac{^{26}{}{C}_2}{^{52}{}{C}_2} + \frac{^{4}{}{C}_2}{^{52}{}{C}_2} - \frac{1}{^{52}{}{C}_2} \right)\]

                                                                = \[\frac{\left( ^{26}{}{C}_2 + ^{4}{}{C}_2 - 1 \right)}{^{52}{}{C}_2}\]

                                                                = \[\frac{\left( 326 + 6 + 1 \right)}{1326} = \frac{330}{1326} = \frac{55}{221}\]

Hence, the required probability is \[\frac{55}{221}\] .