I apologize upfront for any spelling mistakes, I'm not used to writing math in english!

This question was taken from a test done by those willing to undertake a Master's course in Statistics in Federal University of Belo Horizonte (Brazil). I don't have access to its solution so I would like to check with you guys if my resolution seems good and see if you have any other interesting ways of solving it.

"Joseph and Mary play the following game: from a box containing 5 black balls and 2 white balls, they alternately withdraw balls, without replacing them. The winner will be the one who withdraws a white ball first. Determine the probability of Mary winning the game, assuming that she withdraws first, then Joseph, and so on."

I went and called the probability of her winning P, as in:

$$P = {P_1} + {P_2} + {P_3} +...+ {P_n}$$

Where ${P_n}$ stands for the chance of her winning on the n-withdraw. So to win on "round 1", she can pull 2 white balls from a total of 7; to win on her round 2, she needs to miss her first pull, then Joseph has to fail too for her to get it right; and I went on following that thought. It is detailed below:

\eqalign{ & {P_1} = {2 \over 7} \cr & {P_2} = {5 \over 7}{4 \over 6}{2 \over 5} = {4 \over {21}} \cr & {P_3} = {5 \over 7}{4 \over 6}{3 \over 5}{2 \over 4}{2 \over 3} = {{24} \over {252}} \cr & {P_4} = 0 \cr & {P_5} = 0 \cr & ... \cr}

After all that, I got 0,57 or 4/7 as final answer:

$$P = 0,2857 + 0,1905 + 0,0952 \approx 0,57$$

Thank you in advance for any inputs you guys may have. Cheers

• Looks right to me -- so does your English. – saulspatz Jul 3 '18 at 0:51
• Not sure why you got downvoted. Good question with a proposed strategy. +1 – adhg Jul 3 '18 at 3:19

Your answer is correct, here is a way to do it using combinations ...

There are 7 balls, 2 of which are white five are black think of an ordered arrangement of the balls as a 7 character string e.g. "BBWBWB"

There are $\binom 72=21$ equally probable arrangements of balls

To win on the first selection the arrangement must start with "W"

There are $\binom 61=6$ equally probable arrangements like this( you just have to decide where to put the second "W" in the 6 remaining places)

To win on her second selection the arrangement must start with "BBW"

There are $\binom 41=4$ equally probable arrangements like this ( you just have to decide where to put the second "W" in the 4 remaining places)

To win on her thirrd selection the arrangement must start with "BBBBW"

There are $\binom 21=2$ equally probable arrangements like this ( you just have to decide where to put the second "W" in the 2 remaining places)

so $$P(WIN) = \frac{ \binom 61+\binom 41+\binom 21 }{ \binom 72}=\frac{12}{21}$$

• This was great. Thank you very much – Pedro Alonso Jul 3 '18 at 1:59