Question

If you are dealt 5 cards from a shuffled deck of 52​ cards, find the probability of getting two queens and three kings.
The probability is
nothing.
​(Round to six decimal places as​ needed.)

Answer 1

The probability of getting two queens and three kings is $\frac{1}{1082900}$

Solution:

Given that, you are dealt five cards from a shuffled deck of 52 cards  

We have to find the probability of getting two queens and three kings  

Now, we know that, in a deck of 52 cards, we will have 4 queens and 4 kings.

${ probability }=\frac{\text { favarable possibilities }}{\text { number of possibilities }}$

Probability of first queen:

$\text { Probability for } 1^{\text {st }} \text { queen }=\frac{4}{52}=\frac{1}{13}$

Probability of second queen:

$\text { Probability for } 2^{\text {nd }} \text { queen }=\frac{3}{51}=\frac{1}{17}$

Here we used 3 for favourable outcome, since we already drew 1 queen out of 4

And now number of outcomes = 52 – 1 = 51

Probability for first king:

$\text { Probability of } 1^{\text {st }} \text { king }=\frac{4}{50}=\frac{2}{25}$

Here favourable outcomes = 4

And now number of outcomes = 51 – 1 = 50

Probability for second king:

$\text { Probability of second king }=\frac{3}{49}$

Here favourable outcomes = 3, since we already drew 1 king

And now number of outcomes = 50 - 1 = 49

Probability for third king:

$\text { Probability of third king }=\frac{2}{48}=\frac{1}{24}$

Here favourable outcomes = 2, since we already drew 2 king

And now number of outcomes = 49 - 1 = 48

Now the total probability of getting 2 queens and 3 kings from a shuffled deck of cards is:

$=\frac{1}{13} \times \frac{1}{17} \times \frac{2}{25} \times \frac{3}{49} \times \frac{1}{24}=\frac{1}{1082900}$

Hence, the probability is $\frac{1}{1082900}$