Question

Two cards are randomly selected from an ordinary playing deck. What is the probability that they form a blackjack? That is, what is the probability that one of the cards is an ace and the other one is either a ten, a jack, a

queen, or a king?

Short answer

We have$0.048///0.0965$.

Step-by-step solution

Given Information.

Two cards are randomly selected from an ordinary playing deck.

Explanation

$$\begin{gathered} P\left(blackjack\right)=\left[\left(4\mathrm{Cl}{)}^{*}\right(16\mathrm{Cl})\right]/\left(52\mathrm{C}2\right) \\ =(4*16)/\left(1326\right) \\ =0.048 \end{gathered}$$

Explanation.The total number of ten, jack, queen, and king in a deck of cards is equal to 4+4+4+4=16. Choosing 1 card apart from 1 ace out of 16 possible cards is equal to 16C1. Further,  there are  4 aces in a deck of cards. Selecting 1 ace from 4 possible aces is equal to 4C1. Also, there are two positions of a selection of cards which means an ace can be either at the first position or last. That is, the ordering also important.

$$\begin{gathered} P(1aceand1oneiseitheraten,ajack,aqueen,oraqueen)=\left[\left(4\mathrm{Cl}\right)*(16\mathrm{Cl}{)}^{*}2!\right]/\left(52\mathrm{C}2\right) \\ =\left(4162\right)/\left(1326\right) \\ =0.0965 \end{gathered}$$