# probability that you pick three cards from the top and they are from different suits given that the first two you picked are?

You shuffle a normal deck of 52 playing cards.What is the probability that you pick three cards from the top and they are from different suits given that the first two you picked are?

Let us say A is the probability that 3 cards are from different suits.

B: is the probability that first two cards are from different suits.

The answer to this question is as follows: P(A given B)=P(A and B)/P(B)

I know P(B) is:

$$p(\text{two different suits}) = \frac{\dbinom{4}{2}\dbinom{13}{1}^2}{\dbinom{52}{2}}$$

But my problem is that I have no idea how to come up with P(A and B). Thank you guys for your help

• You have chosen 2 different suits for 1st and 2nd draw , now choose one from remaining two suits and pick a card from that suit, hence $^2C_1(^{13}C_1)/^{52}C_1$ – ab123 Sep 18 '17 at 16:28
• yeah right it should be $^2C_1(^{13}C_1)/^{50}C_1$ because you have already chosen 2 cards – ab123 Sep 18 '17 at 16:43

Assuming the first two cards are from distinct suits, there are $50$ cards left in the deck, of which $26$ of them qualify to yield $3$ distinct suits, so the desired conditional probability is $26/50$.
Alternatively, using the formula for conditional probability . . . \begin{align*} P(B) &= \frac{\bigl((4)(13)\bigr)\bigl((3)(13)\bigr)}{(52)(51)}\\[8pt] P(A) &= \frac{\bigl((4)(13)\bigr)\bigl((3)(13)\bigr)\bigl((2)(13)\bigr)}{(52)(51)(50)}\\[8pt] P(A\mid B) &= \frac{P(A \cap B)}{P(B)}\\[4pt] &= \frac{P(A)}{P(B)}\qquad\text{[since $A \subset B$]}\\[4pt] &= \frac{(2)(13)}{50}\\[4pt] &= \frac{26}{50}\\[4pt] \end{align*}