Probability of having1 King and no Queens in 6 cards pick

Question

Lets say we have a deck of 36 cards (4 suits x 9 values).
What is the probability that if you pick 6 random cards there will be at least 1 King and totally no Queens?
I know that the probability of picking 6 cards with no Queens is $36 \choose 32$, however, I have no idea what should I do next

Answer 1

First, you should compute how many "different hands can be dealt". That is fairly easy if you know a bit of combinatorics and this link can give you some inspiration.

$\#\text{possibilities}={{36}\choose{6}}$

Then you have to calculate the number of cases where you have exactly one king and 0 queens. You have 4 suits so you first have to decide one for your king and then you have to take 5 indifferent cards from the 28 remaining (36- 4 queens - 4 kings )

Then you have to calculate the number of cases where you have exactly two kings and 0 queens. And so on.

So the probability is: $$\frac{{{4}\choose{1}}{{28}\choose{5}}}{{{36}\choose{6}}}+\frac{{{4}\choose{2}}{{28}\choose{4}}}{{{36}\choose{6}}}+\frac{{{4}\choose{3}}{{28}\choose{3}}}{{{36}\choose{6}}}+\frac{{{4}\choose{4}}{{28}\choose{2}}}{{{36}\choose{6}}}$$

Answer 2

The requested probability is the following

$$\frac{\binom{4}{1}\binom{28}{5}\binom{4}{0}+\binom{4}{2}\binom{28}{4}\binom{4}{0}+\binom{4}{3}\binom{28}{3}\binom{4}{0}+\binom{4}{4}\binom{28}{2}\binom{4}{0}}{\binom{36}{6}}\approx 27.18\%$$

Answer 3

Hint:

$$ \text{# deals with at least one King and no Queens} \\ = (\text{# deals with no queens})-(\text{# deals with no kings and no queens}) $$ You have already (in the comments) correctly counted that the number of deals with no Queens is $\binom{32}6$ (because a deal with no queens is made by choosing $6$ cards from the $32$ non-queens, without regard to order), so I think you can count the number of deals with no kings and no queens. Once you have done that subtraction, the probability is found by dividing by the total number of deals, $\binom{36}{6}$.