Question

You pick 5 cards out of a deck of 52. What is the probability that you get exactly two spades?

Answer 1

I tried it and thought it was \[\left(\begin{matrix}5 \\ 2\end{matrix}\right)(\frac{ 1 }{ 4 })^2(\frac{ 3 }{ 4 })^3\] But someone else posted somewhere on the internet that it would be \[\frac{ 13 }{ 52 } * \frac{ 12 }{ 51 } * \frac{ 39 }{ 50 } * \frac{ 38 }{ 49 } * \frac{ 37 }{ 48 }\]
But I think now it's: \[\frac{ \left(\begin{matrix}5 \\ 2\end{matrix}\right)(\frac{ 1 }{ 4 })^2(\frac{ 3 }{ 4 })^3 }{ \left(\begin{matrix}52 \\ 5\end{matrix}\right) }\]

Answer 2

there are 13 spades, and 40 non-spades.

so (13C2) (40C3) / (52C5)

Answer 3

oops, i meant 39 non-spade XD

Answer 4

(13C2) (39C3) / (52C5)

Answer 5

Oh that makes sense! thank you!

I have another similar question too, if you don't mind

Answer 6

You pick 5 cards out of a deck of 52. What is the probability that we get at least one card of each suit?

Answer 7

was the solution given? i can work it out but not sure if it's right

Answer 8

No:( I wish lol

Answer 9

we I have (13C1) (13C1) (13C1) (13C1) (48) / (52C5)

each of the first 4 cards are guaranteed to in a different suit. The 5th card can be any card that is remaining

Answer 10

I mean that sounds right to me lol. Thank you!

Answer 11

I'm not sure it's that, but i'm also not sure it's what I have:
\[\frac{ \left(\begin{matrix}4 \\ 1\end{matrix}\right)\left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}48 \\ 1\end{matrix}\right) }{ \left(\begin{matrix}52 \\ 5\end{matrix}\right) }\]
It could be this because you have to choose 1 of 4 suits for each of the first 4 cards, and the last card could be one of 48 other cards that could be picked in any order

Answer 12

Is that how that works?

Answer 13

each suit has 13 right?

Answer 14

13 cards

Answer 15

Yeah, but you're just picking 1 of 4 suits, so I would think that you would just choose 1 of the 4 suits per card and then the last one is one of the remaining 48 cards

but then I got confused because my prof had told me that the n's in the numerator had to add up to equal the n in the denominator (for the n C r) and the same thing for the r's

Answer 16

And any of the cards can be in any order

Answer 17

not sure :/