Question

Suppose that the inside bottom of a box is painted with three colors: 1/3 of the bottom area is red, 1/6 is blue, and 1/2 is white. You toss a tiny pebble into the box without aiming and note the color on which the pebble lands. Then you toss another tiny pebble into the box without aiming and note the color on which that pebble lands. What is the probability that one of the pebbles lands on the color white and the other pebble lands on red? (Enter your probability as a fraction. Hint: How is this exercise different from finding the probability that the first pebble lands on the color white and the second pebble lands on red?)

Answer 1

The probability that pebble lands on red is

P(red) = 1/3

The probability that pebble lands on blue is

P(blue) = 1/6

The probability that pebble lands on white is

P(white) = 1/2

Since more than one pebble can land on same color so it is like sampling with replacement. That is event landing of first pebble is independent of event of landing of second pebble.

So the that one of the pebbles lands on the color white and the other pebble lands on red is

P(first on white and other on red) + P(first on red and other on white) = P(first on white)P(other on red)+P(first on red)P(other on white) = (1/2) * (1/3) + (1/3) * (1/2) = 1/3

Answer: 1/3