What is the probability of drawing a black card then a queen?

Question

My first thought was $(26/52 \times 4/51)$, but then I realized this expression doesn't take into account that the first draw could be a queen as well.

Then I considered $(26/52 \times 3/51)$, but this one doesn't look right.

I also considered the sum of these two $(26/52 \times 4/51) + (26/52 \times 3/51)$, since it considers the two possible cases (if the first draw were or weren't a queen), but I wouldn't know how to defend it further than that.

Answer 1

Just added an Addendum to the end of my answer, to provide a shortcut.

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Consider

$$\left[\frac{24}{52} \times \frac{4}{51}\right] + \left[\frac{2}{52} \times \frac{3}{51}\right]. \tag1 $$

In (1) above, the first term refers to the combined probability of drawing a black non-queen, followed by any queen. The second term refers to the combined probability of drawing a black queen, followed by any queen.

Since the two terms refer to mutually exclusive events (i.e. either the first black card is a queen or it isn't), and since they encompass all of the ways of drawing a black card and then a queen, you can add the two terms to get the desired probability.

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$\underline{\text{Addendum}}$

Shortcut:

Let

• $E_1~$ denote the event that the first card is black.

• $E_2~$ denote the event that the first card is red.

• $E_3~$ denote the event that the second card is a queen.

Then, by symmetrical considerations,

$$p(E_1) = p(E_2), ~~\text{and} ~~p(E_3 ~| ~E_1) = p(E_3 ~| ~E_2).$$

Therefore,

$$\frac{1}{13} = p(E_3) = [ ~p(E_1) \times p(E_3 ~| ~E_1) ~] + [ ~p(E_2) \times p(E_3 ~| ~E_2) ~]$$

$$ = 2 \times [ ~p(E_1) \times p(E_3 ~| ~E_1) ~] \implies $$

$$p(E_1) \times p(E_3 ~| ~E_1) = \frac{1}{13} \times \frac{1}{2}.$$

More intuitively, $~E_1~$ and $~E_3~$ are independent events.

Answer 2

Let $B_i$ be the event of drawing a black card on the $ith$ draw and $Q_i$ the probability of drawing a Queen on the ith draw. Then $$ \begin{align} P(Q_2\mid B_1) &= [P(Q_2\mid B_1 \cap Q_1^c) P(B_1\cap Q_1^c) + P(Q_2\mid B_1 \cap Q_1) P(B_1\cap Q_1)]/P(B_1)\\ &=P(Q_2\mid B_1 \cap Q_1^c)P(Q_1^c\mid B_1) + P(Q_2\mid B_1 \cap Q_1) P(Q_1\mid B_1)\\ &=\frac{4}{51}\frac{24}{26}+\frac{3}{51}\frac{2}{26} \end{align} $$ by the law of total probability and definition of conditional probability. So $$ P(Q_2\cap B_1) = P(Q_2\mid B_1) P(B_1) = \frac{4}{51}\frac{24}{52}+\frac{3}{51}\frac{2}{52} $$ since $P(B_1)=26/52$.