Question

internet is slow for me today ....

Given 10 cards; 3 are red and 7 are blue.

What is the probability of randomly drawing 5 cards and 2 of them are red?

Answer 1

exactly 2 of them are red?

Answer 2

yes

Answer 3

find\[\frac{10!}{5!(10-5)!}\]
which is 252...

out of 252 sets of 5 cards.... how many of them are red?

Answer 4

oops I mean out of those 252 sets how many of them contain exactly 2 reds?

Answer 5

good, we have a total number of outcomes to be 252; of those outcomes, what is the chance of drawing 2 red cards and 3 blue?

Answer 6

lemme do a python simulation

Answer 7

so far we have:
\[\frac{?}{^{10}C_5}\]

Answer 8

so.... we find \[\frac{252!}{2!(252-2)!}\]?

Answer 9

not quite ...

Answer 10

or maybe lol

Answer 11

draw 5 cards and you want exactly 2 red ones right?

Answer 12

252! is a big number :-P

Answer 13

it might be useful to think of it as needing \(^3C_2\) for your successes

Answer 14

yes; draw 5 and exactly 2 are red

Answer 15

we can probably reason it out, and get a nicer answer than this compound fraction, but it would be
\[\frac{\dbinom{3}{2}\dbinom{7}{3}}{\dbinom{10}{5}}\]

Answer 16

yes, a hypergeometric pdf :)

Answer 17

why wouldn;t it be\[\frac{\dbinom{252}{2}}{\dbinom{10}{5}}\]

Answer 18

its a related cousin to the binomial pdf

Answer 19

reasoning for this as you need 2 out of 3 red, 3 out of 7 blue, divided by number of ways to pick 5 from ten

Answer 20

Oh okay

Answer 21

thats cool

Answer 22

how do we do a python simulation to find an answer like this?

Answer 23

10c5 outcomes total

the number of successes would be 3c2 and failures 7c3

Answer 24

binomial pdf relates to probability WITH replacement

the hyper pdf relates to probability WITHOUT replacement

Answer 25

i am not fond of these compound fractions though. we could also reason as follows:
first card red with probability
\[\frac{3}{10}\] second card red give first card red is
\[\frac{2}{9}\] and then 3rd 4th and 5th blue given first stuff is
\[\frac{7}{8}\times \frac{6}{7}\times \frac{5}{6}\]
multiply all this together, and then multiply by number of ways to choose 2 from 5 and you should get the same answer

Answer 26

my calculator says 5/12

Answer 27

in other words (and i will check this number in case i am full of beans) you should also get
\[\dbinom{5}{2}\times \frac{3\times 2\times 7\times 6\times 5}{10\times 9\times 8\times 7\times 6}\]

Answer 28

i got
\[\frac{5}{12}\] for first number, let me try second one

Answer 29

i get 1/24

Answer 30

hmmm

Answer 31

1/24 multiplied by binom(5,2)

Answer 32

5/12 is correct

Answer 33

5/12 is the right one yes

Answer 34

it is
\[\frac{1}{24}\times \dbinom{5}{2}\] what adg said

Answer 35

just as a matter of taste i prefer the "reason it out" method. of course the numbers are the same after you invert, cancel etc, but i am not a fan of compound fractions like
\[\frac{\dbinom{3}{2}\dbinom{7}{3}}{\dbinom{10}{5}}\]

Answer 36

and not to belabor the obvious, but after the onslaught of cancellation this is a fairly simple arithmetic problem

Answer 37

:)