Problem:
Angelina is playing a game which she wins with probability
$0.1$.She must pay
$10$to play, and if she wins, she receives$80$.If Angelina starts out with
$30$, to the nearest thousandth, what is the probability that she wins exactly once before losing it all?
The probability of winning exactly once before losing it all is going to be the probability of her winning any of the first three games then the probability of her losing the next $L$ games which we are yet to define.
Let $L$ be the number of games she loses after winning any of the first three games.
We know that she starts with $30 each game and has both a net profit of $70 and a net loss of $10. This means the total amount of games she can play is going to be:
$$30 + 70 - 10 \cdot L = 0$$
Solving for $L$ we get:
$$L = 10$$
Which means she can play the game 10 + 1 = 11 times before she loses it all.
Now the probability of her winning exactly once is constrained by the fact she needs to win exactly once in the initial three games or else she will lose it all.
We can represent this event as $\mathcal{W}$ and the probability of this event is given by:
$$P(\mathcal{W}) = \binom{3}{1} \cdot 0.1 \cdot 0.9^{2}$$
which give us an answer of approximately 0.243.
After the 3 games we are left with exactly 8 games to play which we can represent as $\mathcal{L}$ and the probability of this event is given by:
$$P(\mathcal{L}) = 0.9^{8}$$
This time we are interested to know the probability of her losing all 8 games and after computing the formula we get about 0.430.
Now we can compute the probability of her winning exactly once before losing it all by:
$$P(\mathcal{W}) \cdot P(\mathcal{L}) = 0.243 \cdot 0.430 = 0.105$$
Which is the probability of her winning exactly once before losing it all.