- #1

stephenranger

- 36

- 1

## Homework Statement

Three friends play a game in which one picks blind–folded from a bag containing white and

black balls. In the bag there are four black balls and one white ball. The player

whose turn it is picks one ball. If the ball is white the player has won; otherwise

the ball is returned to the bag and the next player gets the turn. The turn rotates

until the white ball is picked.

a) What is the probability that the game ends before any of the players has

picked twice?

b) Let the players be A, B, and C, in this order. What is each player’s probability

of winning the game?

## Homework Equations

## The Attempt at a Solution

The following is my solution. But I'm not sure if it is correct. Please correct me if there's any mistake. Thanks.

a)

The probability of the 1st player picks the white ball is P

_{1}= 1/5

The probability of the 1st player picks a black ball and then the 2nd player picks the white ball is P

_{2}= (4/5)x(1/5)

The probability of the 1st player picks a black ball and then the 2nd player picks a black ball and then the 3rd picks the white ball is P

_{3}= (4/5)x(4/5)x(1/5)

So the probability that the game ends before any of the players has picked twice is: P = P

_{1}+P

_{2}+P

_{3}= (1/5) + (4/5)x(1/5) + (4/5)x(4/5)x(1/5) = 61/125 = 0.488

b)

The probability that A picks the white ball is P

_{A}= 1/5

The probability that B picks the white ball is P

_{B}= (4/5)x(1/5)

The probability that C picks the white ball is P

_{C}= (4/5)x(4/5)x(1/5)

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