Probability of a box

Question

A box contains 8 yellow, 5 green and 7 black balls identical to each other except for colour. 3 balls are drawn at random one after the other without replacement. Find the probability that: (i) there will be one ball of each colour. (ii) exactly one ball will be black (iii) at least one ball will be black (iv) the second ball will be black (v) the second ball will be black given that the first is black.

Answer 1

i) 8/20 x 7/19 x 6/18

+ 5/20 x 13/19 x 3/18 + 7/20 x 6/19 x5/18 = probability of 3 yellow or 3 green or 3 black

ii) 7/20 x 13/19 x 12/18 + 13/20 x 6/19 x 12/18 + 12/20 x 11/19 x 8/18 =

probability of exactly one black

iii) calculate the probability of zero black then subtract that from one

to get the probability of at least one black ball

P(no blacks)= 13/20x12/19x11/18

P(at least one black) = 1 - 13/20x12/19x11/18

iv) probability 2nd is black = 7/19

there are 19 balls to draw from, and 7 ways it could be black

v) If the 1st ball is black that leaves 6B, 8Y, and 5G among 19 balls

P(2nd black) = 6/(6+8+5) = 6/19