# Probability of picking 2 queens and 1 king from a deck of cards

Draw 3 cards from deck of 52 cards with replacement. Find probability for drawing 2 queens and 1 king, if we know:
a) at least 1 queen is drawn,
b) at least 2 of the cards are kings, queens, or jacks,
c) all cards are greater than 7.

These are my thoughts:

a) here I think we only draw two cards and probability to draw queen or king is $$\frac{4}{52}$$ so solution should be $$\frac{4}{52}^2$$
b) there are 12 cards to choose from (4 queens, 4 kings and 4 jacks) for the first two draws, third draw is from 52 cards: $$\frac{4}{12}*\frac{4}{12}*\frac{4}{52}$$ c) don't know this one, maybe it's: $${\frac{4}{24}}^3$$

What do you think are correct answers?

• All of your answers are incorrect. The question is asking for conditional probabilities. For instance, the first question is asking for the probability that $2$ queens and $1$ king are drawn given that at least $1$ queen is drawn. Jun 9 '18 at 10:30
• @N.F.Taussig Thank you for help! I will try to learn and find correct solutions. Will edit my post. thanks
– vesa
Jun 9 '18 at 10:56

For all three of these questions, the proper answer depends on applying Bayes' Theorem correctly:

$$P(A|B) * P(B) = P(B|A) * P(A)$$

Where $A$ is the event "draw two queens and a king", and $B$ is the specific known condition in each part. First note that in all three cases, $P(B|A)$ is $1$ since drawing two queens and a king satisfies each given condition. Since the problem asks for $P(A|B)$, we'll use:

$$P(A|B) = \frac{P(A)}{P(B)}$$

So then, what's $P(A)$? (Since that's the same in all three cases)

Well, drawing two queens and then a king does indeed have a probability of

$$\left(\frac{4}{52}\right)^3$$

But "two queens and a king" would imply that the king can be drawn first, second, or third so in fact

$$P(A) = 3 \left(\frac{4}{52}\right)^3$$

So now for the the parts:

In part (a) $B$ is "at least one queen". The chance of "at least one queen" is going to be $1$ minus the chance of "no queens", so:

$$P(A|B) = \frac{3 \left(\frac{4}{52}\right)^3}{1 - \left(\frac{48}{52}\right)^3} = \frac{3}{469}$$

In part (b), $B$ is "at least two face cards". There a few ways to calculate $P(B)$, but I think that the easiest conceptually is to split it into two (mutually exclusive) cases: two face cards and a non-face card ($B_1$), or three face cards.($B_2$)

In the first case, the non-face card can be first, second, or third, so

$$P(B_1) = 3 \left(\frac{12}{52}\right)^2\left(\frac{40}{52}\right) = \frac{270}{2197}$$

And in the second case:

$$P(B_2)= \left(\frac{12}{52}\right)^3 = \frac{27}{2197}$$

So:

$$P(A|B) = \frac{3 \left(\frac{4}{52}\right)^3}{\frac{27}{2197} + \frac{270}{2197}} = \frac{1}{99}$$

In part (c), $B$ is "all cards are greater than 7", so that's saying that every card is 8, 9, 10, J, Q, or K. (I'm assuming here that Ace is counted as lowest card, not highest) so $P(B)$ is just $(24/52)^3$, and

$$P(A|B) = \frac{3 \left(\frac{4}{52}\right)^3}{\left(\frac{24}{52}\right)^3} = \frac{1}{72}$$