What is the probability that the queen of hearts is drawn exactly twice assuming all 3 cards drawn were queens?

Question

3 cards are drawn independently from a standard 52-card deck with replacement. What is the probability that the queen of hearts is drawn exactly twice assuming all 3 cards drawn were queens?

I believe I have to use Bayes rule.

If event A is that a queen of hearts is drawn exactly twice, and event B is the event that all 3 cards drawn were queens, then I would set it up as:

P(A|B) = P(B|A)P(A) / P(B)

P(B|A) is the probability that all 3 cards are queens given a queen of heart is drawn twice. So this would be 1/12? Because the chance of the third card being a queen is just 1/12?

p(A) is the probability that the queen of hearts is drawn exactly twice out of three draws, which is (1/52)(1/52)(51/52).

P(B)...I'm not sure how to compute this.

Answer 1

There are four queens in the deck one of which is the queen of hearts. So what is the probability of a card drawn being the queen of heart given that the card is a queen?

Answer 2

$$P(A|B) = \frac{P(A\cap B)}{P(B)} =\frac{3\times (\frac{1}{52})^2 \times \frac{3}{52}}{(\frac{4}{52})^3}$$

If event A is that a queen of hearts is drawn exactly twice, and event B is the event that all 3 cards drawn were queens.

There are $4$ queens, probability of choosing a queen is $\frac{4}{52}$

Since heart's of queen can come in any order, multiply by $3$.

Answer 3

Compare this to the following question:

I have a bag with only four balls. One of the balls is red, the other three of the balls are green. I draw balls one at a time with replacement and do this a total of three times. What is the probability that we drew the red ball exactly twice?

Take the time to recognize this is exactly the same question as yours after taking care of the conditional probability aspect of the question and will have the same answer as yours. This is an introductory textbook example of a binomial distribution question and has as its final answer:

$$\binom{3}{1}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right) = \frac{9}{64}$$

Answer 4

Let $A$ be the event that a queen of hearts is drawn exactly twice in three draws, and $B$ be the event that all 3 cards drawn were queens.

$$ Then\;\;P(A|B) = \frac{P(A\cap B)}{P(B)} =\frac{\binom32\times (\frac{1}{52})^2 \times (\frac{3}{52})}{(\frac{4}{52})^3}= \frac{9}{64}$$

Answer 5

Alternative approach that avoids Bayes Theorem:

Attack the problem combinatorically, expressing the probability as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}, \tag1 $$

where $D = 4^3.$

So, in (1) above, the denominator of $\left(4^3\right)$ represents the total number of ways that three cards can be drawn from the four card deck of four queens, sampling with replacement.

Here, for convenience the computation of $\left(4^3\right)$ is used for the denominator. However, this contrivance assumes that the order that the cards are drawn is relevant. That is, drawing the Queen of Hearts, and then the Queen of Diamonds is regarded as distinct from drawing the Queen of Diamonds and then drawing the Queen of Hearts.

Therefore, the numerator, $(N)$, must be computed in a consistent manner. From this perspective, how many satisfying ways are there of drawing the three cards?

There are three choices for which of the three cards will not be the Queen of Hearts.

Assume that the first card drawn is not the Queen of Hearts. Then, there are three choices for which other Queen that this first card will be.

Therefore,

$$N = 3 \times 3 = 9.$$

Therefore, the probability is

$$\frac{N}{D} = \frac{9}{64}.$$