# A conditional probability question

1. Aug 22, 2004

In a class of 15 students, 10 are expected to pass maths and 12 are expected to pass english.

how many students are expected to pass maths and english?

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the answer given in the book is 7. i don't understand how this answer was reached. could someone please show me how to calculate this answer.
thanks.

2. Aug 22, 2004

### Zurtex

Assume the 5 people who are expected not to pass maths are expected to pass English. This leaves 7 people who are expected to pass English and pass Maths.

However that is the minimum number it could, all people who are expected to pass Maths could also be expected to pass English. This would mean 10 people could be expected to pass both Maths and English.

So at least 7 people, at most 10 people.

3. Aug 22, 2004

### HallsofIvy

Zurtex is correct. With the given information you can't say how many are expected. Are you sure the problem didn't say "What is the minimum number of students who can be expected to pass both."

4. Aug 22, 2004

it definatly says "how many students are expected to pass maths and english?"

and the only answer given is 7.

that is only part A of the question though, part B is; "show that these two events are not independent"

5. Aug 22, 2004

### robert Ihnot

Maybe the answer was given because it was supposed to be used in part B of the problem, and was not intended to be calculated.

6. Aug 22, 2004

the answer is given in the answers section at the back, it is not in with the question.

7. Aug 22, 2004

### Alkatran

10 out of 15 students are expected to pass maths: 10/15 chance
12 out of 15 students are expected to pass english: 12/15 chance
The chance of passing both is the product of both: 10/15*12/15 =~ 53%
15*.53 =~ 8

It doesn't make sense to assume that any student who doesn't pass math will pass english. This wouldn't take into account very smart or very inattentive students.

Just because it doesn't agree with the book doesn't mean it's wrong.

Oh, and the two aren't independent because of the students. Each student's attributes will affect both tests (sometimes inversely, sometimes the same way).

8. Aug 22, 2004

### needhelpperson

hmmmmmmmmmm

Last edited: Aug 22, 2004
9. Aug 22, 2004

### cepheid

Staff Emeritus
So which answer is correct? I'm curious.

10. Aug 23, 2004

### uart

What if the question asked you to "write down a number between one and ten" and you wrote four but the answer “in the book” was six, who's correct? It's quite simply an incompletely specified problem.

In this case P(A) and P(B) alone just don't fully specify P(A and B) without further assumptions.

In particular,

P(A or B) = P(A) + P(B) - P(A and B)

and also,

P(A and B) = P(A | B) P(B) = P(B | A) P(A)

where P(A | B) reads “Probability of event A occurring given that event B has occurred”.

So just knowing P(A) and P(B) is not enough, you must also know either P(A or B) or one of the two conditional probabilities P(A | B) or P(B | A) in order to fully specify it.

The conditional probably P(A | B), like any probability, can take values between zero and one. The extreme case of P(A | B) = 0 corresponds to “mutually exclusive” events where in this case the occurrence of event B totally precludes event A from occurring. The other extreme case of P(A | B) = 1 corresponds to A and B being so closely related that A must always occur when B occurs (ie, B a subset of A).

In general when P(A | B) has a value that is greater then P(A) then it means that there is some type of positive correlation between event B and event A. Conversely when P(A | B) is less than P(A) then there is some sort of negative correlation, where event B occurring inhibits the chances of event A happening. The other salient position is an intermediate case where P(A | B) is equal to P(A) and thus the occurrence or otherwise of event B has no impact at all on the chances of event A. This is the case of “independent events”.

So in this question there are a whole range of potential answer for P(A and B) depending on what assumption you make about the inter-dependence of the English results and the Maths results.

The extreme cases (that are consistent with the given data) can be found as follows.

1. Clearly P(A and B) <= Min( P(A), P(B) ), so P(A and B) is less than or equal to 10/15.

2. P(A and B) = P(A) + P(B) - P(A or B), whih implies that P(A and B) >= P(A) + P(B) -1.
So in this case P(A and B) >= 10/15 + 12/15 - 1 = 7/15.

Case 1 above (10 students pass both subjects) corresponds to the most optimistic assumption that we can make (while remaining consistent with the given data) regarding the correlation between passing Maths and passing English. Case 2 on the other hand (7 students pass both subjects) corresponds to the most pessimistic (data compatible) assumption and in this scenario we're actually implying that passing English inhibits your chances of passing Maths and visa versa.

The intermediate assumption of independence corresponds to P(A and B) = P(A) P(B) giving a resultant probability of 8/15 (8 students pass both subject).

Possible answers range from 7 through to 10 inclusive, and any answer to this question should have an assumption stated. I'd give zero marks for an answer of 7 without any supporting assumptions.

Last edited: Aug 23, 2004