Question

An urn contains five balls, one marked WIN and four marked LOSE. You and another player take turns selecting a ball at random from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done (Enter your answer in fractions, e.g 1/2) With replacement.

Answer 1

Answer:

The probability that you will win if the sampling is done with replacement is $P(W) = \frac{273}{3125}$

Step-by-step explanation:

If you draw first, you will win if the WIN ball is selected in an odd-turn: 1st, 3rd or 5th.

The probability of getting WIN in the first draw is

$P(X_1=W)=\frac{1}{5}$

that is the probability of getting one specific ball (the WIN one) in the group of five balls.

The probability of getting WIN in the 3rd row, implies that in the first two draws the balls were LOSE:

$P(X_3=W)=P(X_1=L)*P(X_2=L*)*P(X_3=W)=\frac{4}{5}* \frac{4}{5}* \frac{1}{5} =\frac{16}{125}$

The probability of getting WIN in the 5th row, implies that in the first four draws the balls were LOSE:

$P(X_5=W)=P(X_1=L)*P(X_2=L*)*P(X_3=L)*P(X_4=L)*P(X_5=W)\\\\P(X_5=W) =\frac{4}{5}* \frac{4}{5}* \frac{4}{5}* \frac{4}{5}* \frac{1}{5} =\frac{256}{3125}$

The probability that you will win if you draw first is then:

$P(W) = P(X_1=W)+P(X_3=W)+P(X_5=W)= \frac{1}{5} +\frac{16}{125} +\frac{256}{3125} \\\\P(W)=\frac{1+16+256}{3125}= \frac{273}{3125}$