composite probability (with solution)

[go for it]

a year ago i've done wrong composite probability equation which i couldn't fix and i've had another look today and got it solved , it's was fun solving it which to why i shared it

3 red marbles and 4 blue marbles. You put them all in a bag and without looking you pull out three marbles. What is the probability that all three marbles are red?

take your time thinking and then open the spoiler

 Spoiler

my first wrong solution was
you are picking at the first time out of 7
second time out of 6 , third time out of 5
so probability to get it first time 1 red is 1/7
second time is 1/7*6 = 1/42
third time is 1/7*6*5 = 1/210

then someone suggested
( 3/7) *(3/6)*(3/5)=9/210
which is also wrong as you won't have 3 marbles at second and third time

the right answer is
first time 1 red out of 3 red out of 7 total is 3/7
second time is 3*2/7*6 (as there is 2 left with 7 prob. to get one of them * 6 prob to get the red on the first time)
third time is 3*2*1/7*6*5 = 6/210 = 1/35
i've build my first assumption without caring about how many red marbles left which totally screwed up

[CptSky]

You have one chance on the number of permutations of the objects. But, objects are similar, you must not count few permutations several times.

Following the theorem, it's :

 Spoiler

1 / ((3 + 4)! / (3! 4!)) = 1 / 35

[go for it]

Originally Posted by CptSky

You have one chance on the number of permutations of the objects. But, objects are similar, you must not count few permutations several times.

Following the theorem, it's :

 Spoiler

(3 + 4)! / (3! 4!) = 1 / 35

you mean 3! 4! / (3+4)! ?

[CptSky]

Originally Posted by go for it

you mean 3! 4! / (3+4)! ?

No, I missed the inverse when writing The answer is : 1 / ((3 + 4)! / (3! 4!)) = 1 / 35

(3 + 4)! / (3! 4!) will gives you the number of possible permutation where few objects are identical. So, AAABBBB is 1 possibility and not 3 (like A1A2A3BBBB could be). It's the way I would do it.