An urn contains 4 red balls, 2 green balls, and 4 blue balls. If 10 balls are
selected at random with replacement between each draw, what is the probability
that exactly 4 red balls will be selected?
The procedure is to make selection 1, then put it back, then make selection 2,
then put it back, then selection 3, then put it back, then..., selection 10,
then put it back.
There are 4+2+4=10 balls and we are selecting 10 balls.
There are C(10,4) selections where the 4 reds could occur.
Each selection could have been a red ball in 4 ways. That's 44 ways.
Each of the other 6 balls could have been a non-red ball in 6 ways. That's 66 ways.
So the numerator of the probability is = 2508226560
The denominator is gotten by:
(10 ways for the 1st selection) times (10 ways for the 2nd selection) times...
all the way to 10, so the denominator is 1010. = 10000000000
So the answer is 2508226560/10000000000 which by dividing numerator and denominator by 5120,
reduces to 489888/1953125 or about 0.250822656 or rounded off, about 0.25, or
about a fourth of the time.
What is the probability that exactly 2 green balls will be selected?
That's done the same way, only the numbers are different. The denominator will
be the same. You do it! If you have trouble, tell me in the note below.
Edwin