combinatorics - A fair die is thrown until a score of less than 5 is obtained. How to find the probability of less than 3 in the last throw?

A fair die is thrown until a score of less than 5 is obtained. How to find the probability of less than 3 in the last throw?

I am not too sure how to approach this one, any ideas?

Answer

This just says that a throw is less than $5$, that is, $1$, $2$, $3$, or $4$. We are asked for the probability of less than $3$, that is, $1$ or $2$, given that it is $1$, $2$, $3$, or $4$. We should not need theory to see that the probability is $2/4$.

Effectively, the condition "less than $5$" restricts the sample space to $4$ equally likely outcomes.

We can set it up and solve it as a formal conditional probability problem. Let $A$ be the event "less than $3$" and $B$ be the event "less than $5$." We want $P(A|B)$. Since $P(A|B)P(B)=P(A\cap B)$, we need only compute $P(B)$ and $P(A\cap B)$. Easily, the probability that a thrrow is less than $5$ is $4/6$. Also, $P(A\cap B)$ is just $P(A)$, which is $2/6$.

Remark: It is important not to be distracted by the irrelevant detail that there may have been a long string of $5$'s and/or $6$'s before the crucial throw. One could also do the computation by taking into account these irrelevant throws. More work, with result, when the smoke clears, $1/2$.