# Three cards are drawn from a standard deck of 52 cards. What is the possibility that a jack, a queen, and a king are selected, drawn in succession without?

Mar 28, 2017

$\frac{8}{15675.}$

#### Explanation:

Let, $J =$ the Event that a Jack has been selected from a std.

deck.

$Q \mathmr{and} K$ denote similar Events for Queen and King resp.

A std. deck have a total of $52$ cards in which, $4$ are Jacks.

Hence, Prob. of drawing a Jack in the first run, is given by,

$P \left(J\right) = \frac{4}{52.} \ldots \ldots . \left(1\right) .$

Now, for the event $Q ,$ the Jack card drawn while occurence of

$J$ is not to be replaced back in the deck, the deck is left with

$52 - 1 = 51$ cards with $4$ Queens in it.

Therefore, the Prob. of getting a Quuen in the second draw, having

known that the event $J$ has occurred, is given by,

$P \left(\frac{Q}{J}\right) = \frac{4}{51.} \ldots \ldots \ldots . \left(2\right) .$

On a similar arguement, we find, $P \left(\frac{K}{Q \cap J}\right) = \frac{4}{50.} \ldots \ldots . . \left(3\right) .$

Now, the Reqd. Prob.$= P \left(J \cap Q \cap K\right)$

$= P \left(J\right) P \left(\frac{Q}{J}\right) P \left(\frac{K}{Q \cap J}\right) = \left(\frac{4}{52}\right) \left(\frac{4}{51}\right) \left(\frac{4}{50}\right) = \frac{8}{15675.}$

Enjoy Maths.!