# Thread: Conditional probability theorem (possiblying involving Bayes')

1. ## Conditional probability theorem (possiblying involving Bayes')

A student either knows the answer or guesses on a multiple choice test.

Let $\displaystyle p$ be the probability the student knows the answer and $\displaystyle 1-p$ be the probability the student guesses.

Assume a student that guesses will be right with a probability of $\displaystyle 1/m$. (where $\displaystyle m$ is the number of multiple choice alternatives.

What is the probability that the student knew the answer to a question given that it was answered correctly?

I am pretty sure this problem uses Bayes Thm, but I am not sure where to begin. Any help please? Thank you.

-Jame

2. Originally Posted by Jame
A student either knows the answer or guesses on a multiple choice test.

Let $\displaystyle p$ be the probability the student knows the answer and $\displaystyle 1-p$ be the probability the student guesses.

Assume a student that guesses will be right with a probability of $\displaystyle 1/m$. (where $\displaystyle m$ is the number of multiple choice alternatives.

What is the probability that the student knew the answer to a question given that it was answered correctly?

I am pretty sure this problem uses Bayes Thm, but I am not sure where to begin. Any help please? Thank you.

-Jame
Draw a tree diagram and use it to insert the appropriate probabilities into the conditional probability formula. I get

$\displaystyle \displaystyle \frac{.....}{..... + \frac{1-p}{m}} = ....$

where the ..... are for you to fill in.

I partitioned the sample space into the events
$\displaystyle K$ - knowing the answering
$\displaystyle G$ - guessing the answering

And I also have $\displaystyle C$, the event where the answer is correct
$\displaystyle C$ has probability $\displaystyle 1$ if you know the answer
$\displaystyle C$ has probability $\displaystyle 1/m$ if you guess

Therefore $\displaystyle P(K|C) = \frac{P(K \cap C)}{P(C)} = \frac{1*p}{1*p + \frac{1-p}{m}}$

Is this correct?

4. Originally Posted by Jame

I partitioned the sample space into the events
$\displaystyle K$ - knowing the answering
$\displaystyle G$ - guessing the answering

And I also have $\displaystyle C$, the event where the answer is correct
$\displaystyle C$ has probability $\displaystyle 1$ if you know the answer
$\displaystyle C$ has probability $\displaystyle 1/m$ if you guess

Therefore $\displaystyle P(K|C) = \frac{P(K \cap C)}{P(C)} = \frac{1*p}{1*p + \frac{1-p}{m}}$

Is this correct?
Yes.