Two cards are chosen at random from a deck of 52 playing cards. what is the probability that they

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There are

cards to start.
Four of these cards are kings.
Therefore the probability of drawing a on the first drawing is or .
After drawing king there are only cards remaining, of which are kings. The probability of drawing a on the second drawing is .
The probability of drawing kings a is therefore the of both draws or

=
= which reduces to

=

so, your answer is b)

Option 1 : \(\frac{1}{{17}}\)

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Concept

\({\bf{C}}\left( {{\bf{n}},\;{\bf{r}}} \right) = \frac{{{\bf{n}}!}}{{{\bf{r}}!\left( {{\bf{n}} - {\bf{r}}} \right)!}}\)

Calculation

From 52 cards, 2 can be chosen in C(52, 2) ways.

There are 4 suits in a deck of 52 cards and each number is present in all the 4 suits.

There are 13 numbers form 1(Ace card) to 13(King card).

If we consider the number 1, then from 4 Aces, 2 can be chosen in C(4, 2) ways.

There are 13 numbers in total so, total ways in which two same numbers can be chosen is 13 × C(4, 2)

Required probability

\( = \frac{{13\; \times \;{\rm{C}}\left( {4,{\rm{\;}}2} \right)}}{{{\rm{C}}\left( {52,{\rm{\;}}2} \right)}}\)

\(= \frac{{13\; \times \;12}}{{52\; \times \;51}}\)

\(= \frac{1}{{17}}\)

Correct option is (1).

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