Question

A bag contains 7 red balls and 3 white balls. Andrew,Betsy, Cam and Doug, in that order. are going to randomlypick a ball out of the bag and not replace it. The firstperson to draw a white ball wins SI60. What is theprobability that Doug will win the S160?

Answer 1

We have the following events:

$\begin{gathered} R_1=\text{Andrew picks a red ball} \\ R_2=\text{Betsy picks a red ball} \\ R_3=\text{Cam picks a red ball} \\ R_4=\text{Doug picks a white ball} \end{gathered}$

since the game is without replacement, then the probabilites are:

$\begin{gathered} P(R_1)=\frac{7}{10} \\ P(R_2)=\frac{6}{9} \\ P(R_3)=\frac{5}{8} \\ P(R_4)=\frac{3}{7} \end{gathered}$

therefore, if we calculate the probability of the intersection of these events (since we want them to happen at the same time) we get:

$P(R_1\cap R_2\cap R_3\cap R_4)=\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}\cdot\frac{3}{7}=\frac{630}{5040}=\frac{1}{8}_{}$

Therefore, the probability that Doug win the $160 is 1/8