Two cards are drawn from a well shuffled pack find the probability that both are kings

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 \[\frac{55}{221}\] .

Text Solution

Solution : Two cards are drawn from a well shuffled deck of cards. `n(S)=(52_(C_2))` A be the event of getting black cards `n(A)=(26_(C_2) ) ` `P(A)=(26_(C_2) /(52_(C_2)))` `=(26×25)/(52×51)` B be the event of getting both king cards `P(B)=(2_(C_2)/(52_(C_2)))=(4×3)/(52×51) ` Also, `P(A∩B)=(2_(C_2)/(52_(C_2)))` `=(2×1)/(52×51 )` Now, `P(A∪B)=P(A)+P(B)−P(A∩B)` `=(26×25)/(52×51)+(4×3)/(52×51)−2/(52×51)` `= (660)/(52×51)=(55)/(221)`