Question

Problem 5. (20 pts) Consider three boxes and 12 balls, and exactly three of the balls are red. The first box fits any three balls, the second box fits any four balls and the third box will fit any five balls. Let us randomly put the balls into the boxes. Show that the probability that all three red balls end up in the same box is 6.81%.

Answer 1

number of ways to distibute 12 balls in 3 boxes =N(select 3 balls from 12 for first box then select 4 balls from remaing 9 for 2nd box and then select 5 balls from 5 for last box)

=¹²C₃*⁹C₄*⁵C₅=220*126*1 =27720

number of ways all red balls in box 1 =N(select 4 balls from remaing 9 for 2nd box and then select 5 balls from 5 for last box=⁹C₄*⁵C₅=126*1=126

number of ways all red balls in box 2 =N(select 1 balls from remaing 9 for 2nd box and then select 5 balls from 8 for last box)=⁹C₁*⁸C₅=9*56=504

number of ways all red balls in box 3 =N(select 2 balls from remaing 9 for 3rd box and then select 5 balls from 7 for 2nd box)=⁹C₂*⁷C₄=36*35=1260

hence P(all red balls in same box) =(126+504+1260)*100/27720 =6.81%