Probability of winning a doll

Question

A man pays 1 dollar for a throw to try to win a 3 dollars doll.

His probability of winning on each throw is 0.1. Find

1. Probability that two throws will be required to win the doll?

2. Probability that more than three throws will be required to win the doll?

3. The expected number of throws needed to win the doll?

My observation is that the probability function must be a binomial but it's not correct if I use this approach so I will need guidance on how to get the answers to this question. If it must be a binomial I needed the number of trials to be given in the question but it is the case here.

Answer 1

Since you ask for guidance and not a plain answer, I suggest you read about the geometric distribution: https://en.wikipedia.org/wiki/Geometric_distribution

Answer 2

Let $p$ be the probability of winning the doll in a throw.

1. If two throws are required, it means that the first throw failed and the second succeeded. The probability of failure is $1-p$, so the total probability of this event (winning the doll after two throws) is $p(1-p)=0.1\cdot 0.9=0.09$.
2. If more than three throws is required, this means there must have been three failures, giving $(1-p)^3=0.9^3$.
3. The number of throws before winning the doll is expected to be $1/p=10$. This makes intuitive sense, and is even given by a normal way of stating a probability: "One in ten."

Note that in the binomial distribution, we want to find the probability of $k$ successes in $N$ trials and there, the order of the outcomes doesn't matter (hence the binomial coefficient). But for this problem, order matters very much, so what is needed is instead a geometric distribution.