Three gamblers, A, B and C, take 12 balls of which 4 arewhite and 8 black. They
ID: 2953018 • Letter: T
Question
Three gamblers, A, B and C, take 12 balls of which 4 arewhite and 8 black. They play with the rules that the drawer isblindfolded, A is to draw first, then B and then C, the winner tobe the one who first draws a white ball. What is the ratio of theirchances? Cananyone help with this. I know I start a treewhere A-----1/3--------W = = =---- 2/3-------B I'mnot sure where to go from here to solve this littleriddle. Three gamblers, A, B and C, take 12 balls of which 4 arewhite and 8 black. They play with the rules that the drawer isblindfolded, A is to draw first, then B and then C, the winner tobe the one who first draws a white ball. What is the ratio of theirchances? Cananyone help with this. I know I start a treewhere A-----1/3--------W = = =---- 2/3-------B I'mnot sure where to go from here to solve this littleriddle. Cananyone help with this. I know I start a treewhere A-----1/3--------W = = =---- 2/3-------B I'mnot sure where to go from here to solve this littleriddle.Explanation / Answer
I believe the first guy has a 1/3 chance. The second guy has 2/9 chance, as there is a 2/3 chance he gets to draw a ball and 1/3 chance it's a white one. The third guy has a 4/27, as there is 2/3 chance of first guy getting the wrong one and 2/3 for the second as well. so 2/3 x 2/3 x 1/3.
I think at least