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)

Vinu P.

what is the calculation behind binom.dist(3,6,2/3,1)?10/30/20