Ina box contain 21 marbles: 6 white, 7 black, and 8 yellow. Randomly pick out one marble and return it back to the box for 8 times. What is the possibility to get exact 4 white and 3 black marbles?

Question

The answer has four decimal place.

Answer 1

probability of selecting a white is 6/21 = 2/7 = 0.286

probability of selecting a black is 7/21 = 3/7 = 0.429

probability of selecting a yellow is 8/21 = 0.381

select 4 white, 3 black and 1 yellow marble

8!

______ (0.286)^4 (0.429)^3 (0.381)^1

4!3!1!

Answer 2

Because we put the ball back, you just need to use the multinomial distribution rather than combinatorics.

Thus, P(white) = 6/21 = 2/7

P(black) = 7/21 = 1/3

P(yellow) = 8/21

Since we draw 8, and 4 white and 3 black, there must be 1 yellow.

Then, P(x1 white, x2 black, x3 yellow) =

n!/x1!x2!x3! p1^x1 p2^x2 p3^x3 where n = x1 + x2 + x3

and

P(4 white, 3 black, 1 yellow) =

8!/4!3!1!*(2/7)^4*(1/3)^3*(8/21)^1 =

2^7*7*3^2*5/(2^4*3^2)*2^7/(7^5*3^4) =

2^10*5/(7^4*3^4) =

5120/194481 =

0.0263264791933402