Probabilty Question: Color of the Marble

Question

Mr.W puts two blue marbles and one green marble into his pocket. Then he randomly takes out one. After recording the color of the marble, he puts the marble back. He repeats this 6 times.

a.What is the probability that he has no more than three blue marbles?

b. What is the probability that he has no more than three blue marbles in a row?

c.What is the probability that he has no two blue marbles in a row?

Answer 1

We assume with the repeat 6 times statement that he draws a total of 6 balls and not 7.

This is sampling with replacement, so samples are independent, and P(blue) = 2/3

a P(no more than 3 blue marbles) = P(3 or less blue marbles) = from Excel, binom.dist(3,6,2/3,1) = .319616 or, for the exact probability, multiply by 3^6 for the numerator and denominator, and we get 233/729

b) This is 1 - P(4 in a row)

Let A = marbles 1, 2, 3, 4

B = marbles 2, 3, 4, 5

C = marbles 3, 4, 5, 6

P(4 in a row) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC) =

(2/3)^4 + (2/3)^4 + (2/3)^4 - (2/3)^5 - (2/3)^5 - (2/3)^6 + (2/3)^6 = 3*(2/3)^4 - 2*(2/3)^5 = 16/27 - 64/243 = 80/243

(Note that P(AC) does not equal P(AB) and P(BC), as AC requires 6 blue marbles, while the other two require 5)

Then, 1 - 80/243 = 163/243 = .670782

c)P(no 2 blue marbles in a row) = P(0 blue marbles) + P(1 blue marble) + P(2 blue marbles not in a row) + P(3 blue marbles not in a row) =

(1/3)^6 + C(6,1)*(1/3)^5*(2/3) + C(5,2)*(1/3)^4*(2/3)^2 + C(4,3)*(1/3)^3*(2/3)^3 =

1/729(1 +6*2+10*4+4*8) = 85/729 = .116598

(Note that on the combinations, the first number decreases; this handles the fact that the blue marbles can't be together)