2
$\begingroup$

Two cards are drawn at random from a standard deck of 52 cards, without replacement. What is the probability that both cards are drawn are queens?

Its a permutation cause the order matter, I mean its a pack of cards and cards have order so its going to be different if you choose anything else right? I'm just guessing not sure.

So the $$ P(Queens) = ? / 52P2 $$ Whats going to be the numerator? I mean i said the denominator was 52P2 because its the total number of outcomes and you pick 2 out of it?

$\endgroup$
5
$\begingroup$

When order is important it is in the selection not what is selected from. The question only wants you to draw two queens. It did not say you had to draw the queen of clubs and then the queen of hearts, just any two queens.

What is the probability of drawing the first queen? Well there are 4 queens in the shuffled deck off 52 cards so the probability is $\frac{4}{52}$

What is the probability of drawing the second queen? This time there are 3 queens left and only 51 cards left in the deck: $\frac{3}{51}$

The probability of these events occurring is therefore: $$\frac{4}{52}\times\frac{3}{51}=\frac{1}{221}$$

$\endgroup$
5
$\begingroup$

If you want to use "combinations," note that there are $\binom{52}{2}$ ways to draw a two-card hand. All these hands are equally likely.

Now we count the "favourables," the number of two-queen hands. There are $\binom{4}{2}$ such hands. It follows that the probability of a two queen hand is $\dfrac{\binom{4}{2}}{\binom{52}{2}}$.

Alternately, imagine drawing the cards one at a time. Record the result as an ordered pair $(a,b)$, where $a$ is the first card drawn, and $b$ is the second card drawn. Then there are $(52)(51)$ possible outcomes, and they are all equally likely.

Now count the number of ordered pairs consisting of two queens. There are $(4)(3)$ of these. So the probability of two queens is $\dfrac{(4)(3)}{(52)(51)}$.

$\endgroup$
2
$\begingroup$

$P(\text{both queens})= P(\text{1st card is queen})\cdot P(\text{2nd card is queen |1st card is queen})=\dfrac{4}{52}\cdot \dfrac{3}{51}$

$\endgroup$
  • $\begingroup$ Why could you explain ? $\endgroup$ – MATH ASKER Dec 16 '15 at 0:56
2
$\begingroup$

We have $\dfrac{4}{52}\cdot \dfrac{3}{51}$ is the answer.

$\endgroup$
  • $\begingroup$ Why could you explain? $\endgroup$ – MATH ASKER Dec 16 '15 at 0:55
  • $\begingroup$ There are $4$ queens in a deck of cards and $52$ cards total. Thus, drawing without replacement yields that answer. $\endgroup$ – John Ryan Dec 16 '15 at 0:57
1
$\begingroup$

Like many combinatorics problems, there are all kinds of methods to answer this:

$$\begin{align*}\text{P(both queens)} &= \text{P(first card queen)} \cdot \text{P(2nd card queen | first card queen)} \\ &= \frac{4}{52} \cdot \frac{3}{51} \\ &= \frac{12}{2652} = \frac{3}{663} \end{align*} $$

There are $4\cdot 3 = 12 $ permutations (order matters) of the first two cards where both are queens. There are $52 \cdot 51 = 2652$ permutations of the first two cards. Each permutation is equally likely. Probability both are queens is then $\frac{12}{2652} = \frac{3}{663}$

There are $\frac{4 \cdot 3}{2} = 6$ combinations (order doesn't matter) of the first two cards where both are queens. There are $\frac{52 \cdot 51}{2} = 1326 $ combinations of the first two cards. Each combination is equally likely. Probability both are queens is then $\frac{6}{1326} = \frac{3}{663}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.