The Fair Problem

• At the beginning, Joan obtains three balls paying 2$

• Joan could obtain three extra balls paying 4$, 8$ and 16$

Notice that he can't obtain more than 6 balls because he hasn't enough money.

Therefore, Joan has a maximum of 6 throws to try to introduce 3 balls into the hole. We are going to define this random variable:

X: Number of throws Joan needs until he inserts three balls into the hole (X=3,4,5,6)

We are interested to calculate:

$P(3\leq X \leq6)$ $=$ $P(X=3)$ $+$ $P(X=4)$ $+$ $P(X=5)$ $+$ $P(X=6)$

$P(X=3)$ $=$ $0.35^3 = 0.042875$

$P(X=4)$ $=$ $[{4\choose3}-{3\choose3}] \cdot 0.35^3 \cdot 0.65 = 3 \cdot 0.35^3 \cdot 0.65 = 0.08360625$

$P(X=5)$ $=$ $[{5\choose3}-{4\choose3}] \cdot 0.35^3 \cdot 0.65^2 = 6 \cdot 0.35^3 \cdot 0.65^2 = 0.108688125$

$P(X=6)$ $=$ $[{6\choose3}-{5\choose3}] \cdot 0.35^3 \cdot 0.65^3 = 10 \cdot 0.35^3 \cdot 0.65^3 = 0.1177454688$

Adding all this probabilities we obtain our final answer

$P(3\leq X \leq6)$ $=$ $0.042875 + 0.08360625 + 0.108688125 + 0.1177454688 = \fbox{0.3529148438}$