Question

An urn contains 3 balls, 1 of which is red, 1 of which is yellow, and 1 of which is blue. Four balls are drawn with replacement from the urn. What is the probability that all 3 balls are seen in these 4 draws?

Answer 1

total number of ways to select 4 balls =3⁴ =81 (as there are 3 choices for each draw)

number of ways to draw so that each ball is seen =N(select ball which will be chosen twce and then distributed 2 same color ball and 1 of each different color ball)=³C₁*4!/(2!*1!*1!)=3*12 =36

probability that all 3 balls are seen in these 4 draws =36/81=4/9