1. ## please help me working out the probablity for this?

The question is: A student is chosen at random, find the probability that they

- do not attend the revision sessions but improve their grade. (which i have the answer as 0.04 and it is correct. )

- improve their grade (answer = 0.85)

Now, with that given information i need to do this:

A student improves their grade, find the probability that they attended the revision sessions.
I'm not sure how to do this, but the answer is 0.95 - please show me the steps for this so i can do the others.

thank you

2. ## Re: please help me working out the probablity for this?

With just the information you give, there s no way to answer that question- and you were surely given additional information. How did you get the "0.04" and "0.85"?

3. ## Re: please help me working out the probablity for this?

oh gosh sorry - yes i thought i posted it but the second half didnt come up in my question.

anyway a tree diagram was included, here is the link to it:
Welcome to Flickr!

thanks for your help

4. ## Re: please help me working out the probablity for this?

Originally Posted by emily9843
The question is: A student is chosen at random, find the probability that they

- do not attend the revision sessions but improve their grade. (which i have the answer as 0.04 and it is correct. )

- improve their grade (answer = 0.85)

Now, with that given information i need to do this:

A student improves their grade, find the probability that they attended the revision sessions.
I'm not sure how to do this, but the answer is 0.95 - please show me the steps for this so i can do the others.

thank you
Originally Posted by emily9843
oh gosh sorry - yes i thought i posted it but the second half didnt come up in my question.

anyway a tree diagram was included, here is the link to it:

thanks for your help
Hi emily9843!

You are asked for a conditional probability.
The probability on A given that B is true is:

$P(A | B) = \frac {P(A \wedge B)} {P(B)}$

In your case that is the chance that they attended, given that they improved, which is:

\begin{aligned}P(attended|improved) &= \frac{P(attended \text{ and } improved)}{P(improved)} \\ &= \frac {0.9 \cdot 0.9} {0.9 \cdot 0.9 + 0.1 \cdot 0.4} = \frac{0.81}{0.85} = 0.95 \end{aligned}