College probability question.

punk_one82

Before I start, this isn't homework or anything. I'm studying for my upcoming exams and have nowhere to get answers from right now. I've done the question and just looking to see if someone can verify my answer or give their answer or whatever.

The question :
At the end of an examination paper students were asked to tick one of two boxes indicating whether or not they believed they would score about the average mark for the class on that examination. After the examination papers were graded it was found that while 70% of students did indeed score about the average, only 60% of the class indicated the belief that they would do so. However, of those who did in fact score about the class average it transpired that 80% indicated the belief that they would do so. If a student is picked at random from this class find the probability that they indicated the belied that they would score above average if in fact they did not do so.
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The answer I got was 0.404 or 101/250. I think I may have gone wrong somewhere. Anyone tell me the answer they get and how they went about doing it? Cheers for the help in advance.

Fremen

Let A be the event that a randomly chosen student believed he would score above average.

Let B be the event that a randomly chosen student actually scored below average.

We're looking for P(A|B).

However, of those who did in fact score about the class average it transpired that 80% indicated the belief that they would do so.

This gives us P(A |B-compliment)

I guess you can solve it from here.

MathsManiac

You used the term "about the average" mostly, but then "above the average" towards the end. Was that deliberate?

If the question consistently referred only to "about the average", then one could survive without a definition of what that means, by just assuming it's consistently interpreted. However, if you intend to also refer to "above average", then you need to give further information about the distribution of scores and beliefs. At a minimum, you would need to state how the people who were not "about the average" are divided between "below average" and "above average", and give the corresponding information about their beliefs.

(By the way, it's not clear to me precisely how one could use the ticking of one of two boxes to separate students into more than two belief categories. Two boxes would give yes/no to the question "do you think you will score about the average", but that would give you no information about whether they believed they would be above or below.)

punk_one82

Sorry for any confusion, I was typing that in a rush. It was supposed to be above, not about.

MathsManiac

In that case, here's another way of looking at it: pretend there are 100 people in the class. Draw up a 2-way table, with the labels "above average" & "not above average" across the top, and "predicts above average" and "doesn't predict above average" on the left. Try to fill in the numbers in the four cells of the table in such a way as to give the correct row and column totals.

After the table is complete, it should be easy enough to read off the required conditional probability.

hivizman

Interesting that 70% of students score "above average". That implies quite a skewed distribution of marks (I'm assuming that "average" here implies "mean" rather than "mode" - it can't imply "median" by definition). In examinations, a small number of extremely low marks when most students get high marks can lead to such a skewed distribution.

The other thought is that this is an attempt to satisfy the sort of pious but innumerate Education Minister who says "no child should achieve a mark that's below average".

MathsManiac

Indeed. And isn't it terrible that after all the investment made in special needs education, one in every twenty children is still below the fifth percentile!

punk_one82

Any chance someone can post an answer to this? It's irritating me that I've nothing to compare my answer to.

MathsManiac

2/15.

Method 1: Using the technique I suggested earlier:
The total of the first column is 70, so the total of the second column is 30.
The total of the first row is 60, so the total of the second row is 40.
Of the 70 in the first column, 80% = 56 people are in the first cell.
The remaining three cells follow easily:

56 04 | 60
14 26 | 40

---

70 30 | 100

The question now becomes: given that you've picked someone in the second column, what is the probability that they are in the top cell? Ans: 4/30.

Method 2: Using technique and laballing of events as suggested by Fremen:
The question gives P(A) = 0.6
The question gives P(B') = 0.7
Therefore, P(A') = 1-P(A) = 0.4 and P(B) = 1-P(B') = 0.3
The question gives P(A|B') = 0.8
We want P(A|B).
Not quite sure whether Fremen intended to complete as follows, but here goes:
P(A|B)P(B) + P(A|B')P(B') = P(A)
So P(A|B)(0.3) + (0.8)(0.7) = 0.6
Solve to get P(A|B) = 2/15.

Howzat?

punk_one82

Thanks for that. Really needed something to compare mine to.

Fremen

Yep MM, I was thinking of using the law of total probability.