I have the following exercise:
There are n marbles in a bag, 2 marbles are blue. Arthur plays a game, in which he randomly takes marbles, one after another, without replacement. The game is over when he draws a blue marble.
a) Find the probability in terms of n that the game ends on second draw.
b) Let n= 5. Find the probability that the game will end on fourth draw.
Arthur plays the game when n = 5. He pays \$20 to play and can earn money back depending on the number of draws it takes to obtain a blue marble. He earns no money back if he obtains a blue marble on the 1st draw. Let M be the amount of money that he earns back playing the game. If numbe of draws is 1, he gets \$0; if it is 2, he gets \$20; if it is 3, he gets 8k dollars; and if it is 4, he gets 12k dollars (k is the constant).
c) Find k so that it is a fair game.
The results that I got are:
a) $\frac{n-2}{n} \frac{2}{n-1}$
b) $\frac{3}{5} \frac{2}{4}\frac{1}{3} \frac{2}{2} = \frac{1}{10} $
c) $1 \cdot 0 + 2 \cdot 20 +3 \cdot 8k + 4 \cdot 12k - 20 = 0 $ what implies k= $\frac{ - 5}{6}$.
Could someone check whether my results are correct? Thanks in advance.
The computation for the third part is not correct.
If we let $B$ be the number of times until we draw a blue marble. It should be
$$0\cdot Pr(B=1) +20\cdot Pr(B=2)+8k\cdot Pr(B=3)+12k\cdot Pr(B=4)-20=0 $$
That is first, we have to compute the pmf of $B$ to find $k$.