Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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, $1/2$.

share|improve this answer
    
Thanks Andre, You make it look so simple :) In my book there was a solution involving summation of infinite Geometric progression. –  Quixotic May 4 '12 at 15:33
2  
It is useful to think about what the problem really is about, instead of thinking about what tools the wording reminds you of. –  André Nicolas May 4 '12 at 15:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.