Problem Solving

A bag has 4 red, 5 blue and 2 green marbles. If 5 marbles ( 2 red, 2 blue & 1 green ) are selected one after the another with out replacement, what is the probability of drawing those 5 marbles?
Please solve for the probability of each marble and whats the probability if the marbles are replaced?

(2 red, 2 blue and 1 green) from (4 red, 5 blue and 2 green)

Without replacement:
4C2/11C2 * 5C2/9C2 * 2C1/7C1

With replacement:
4C2/11C2 * 5C2/11C2 * 2C1/11C1

hey,
and how would you solve it, if the order would matter?

Thanks!

[email protected] wrote:hey,
and how would you solve it, if the order would matter?

Thanks!

Result does not change even if the order changes. see here:
Let us change the order now.
(1 green,2 red, 2 blue) from (2 green,4 red, 5 blue)

Without replacement:
2C1/11C1 * 4C2/10C2 * 5C2/8C2 = 2/11 * 4*3/10*9 * 5*4/8*7 = (2*4*3*5*4)/(11*10*9*8*7)
The other order considered in my earlier explanation is: 4C2/11C2 * 5C2/9C2 * 2C1/7C1
= (4*3*5*4*2)/(11*10*9*8*7)
Both are equal

With replacement:
2C1/11C1 * 4C2/11C2 * 5C2/11C2 = (2*4*3*5*4)/(11*11*10*11*10)
The other order considered in my earlier explanation is: 4C2/11C2 * 5C2/11C2 * 2C1/11C1
= (2*4*3*5*4)/(11*11*10*11*10)
Both are equal

its differs from the official answer, which is 20/77
got from 4c2*5c2*2c1/11c5 = 20/77

please explain why the answer differs when you consider the probability of each marble separately?

kksekar wrote:its differs from the official answer, which is 20/77
got from 4c2*5c2*2c1/11c5 = 20/77

please explain why the answer differs when you consider the probability of each marble separately?

What is the OA for the case "with replacement"

with replacement is not part of the Official question and so there is no OA