Probability of a student failing a prelim but passing the course

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

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The % of course passing students who fail the prelim test is 38%. So, the required probability is 38 %.
Isn't the probability of a student failing the prelim when he his passing the course is known already = probability of a student who failed the prelim but pass the course?

Another idea is to calculate total no. of students who failed the prelim let's say, N(pre F).
Then the required probability is $\frac {\text{no. of students who have passed the course but failed the prelim i.e. N(course P, prelim F)} }{\text{ non. of students who failed the prelim i.e.N( pre F)}$

Let's take the total no. of students to be 100.N( pre F) = 100 *95%*4% +100 * 5% * 75%= 7.55
N(course P, pre F) = 100 *95%*4%=3.8
Thus, the req. probability = 3.8/7.55=0.503

I have solved the question, but I don't understand what I had done here.

 

[Ray Vickson]

Homework Statement

View attachment 215217

Homework Equations

The Attempt at a Solution

View attachment 215218 [/B]
View attachment 215218

The % of course passing students who fail the prelim test is 38%. So, the required probability is 38 %.
Isn't the probability of a student failing the prelim when he his passing the course is known already = probability of a student who failed the prelim but pass the course?

Another idea is to calculate total no. of students who failed the prelim let's say, N(pre F).
Then the required probability is $\frac {\text{no. of students who have passed the course but failed the prelim i.e. N(course P, prelim F)} }{\text{ non. of students who failed the prelim i.e.N( pre F)}$

Let's take the total no. of students to be 100.N( pre F) = 100 *95%*4% +100 * 5% * 75%= 7.55
N(course P, pre F) = 100 *95%*4%=3.8
Thus, the req. probability = 3.8/7.55=0.503

I have solved the question, but I don't understand what I had done here.

(1) This question does not belong in the Introductory Physics Forum; it belongs in Precalclulus Mathematics.
(2) How can you claim to have solved the problem when you do not know or understand what you did? That makes no sense!
(3) The way to deal with such problems is through the use of conditional probabilities: $P(A|B)$ = probability of $A$, given that $B$ occurs. You are told that among the students who pass the course, 96% of them passed the prelim; that is you are told that P(pass prelim|pass course) = 0.96. You are also told that among the students who fail the course, 25% passed the prelim; that is, P(pass prelim|fail course) = 0.25. Finally, you are also told that P(pass course) = 0.95. From those givens you are asked to find P(pass course|fail prelim). That is a standard exercise in Bayes' Theorem.

One way to deal with such problem if you don't know Bayes' theorem, is to do a tabular method like the one you attempted. However, you need to break down cases differently: we can have (pass course, pass prelim), (pass course, fail prelim), (fail course, pass prelim) and (fail course, fail prelim).

Suppose we take $N = 1000$ students. Among these, $N_p = (0.95)(1000) = 950$ pass the course and $N_f = 1000-950 = 50$ fail the course. So far, we have obtained the following partial table:
$$\begin{array}{r|cc|r}
& \text{pass prelim}&\text{fail prelim} & \text{Total} \\ \hline
\text{pass course} & ? &? & 950 \\
\text{fail course} &? & ?& 50\\ \hline
\text{Total} & ? & ? & 1000
\end{array}
$$
Let us fill in the missing entries in the first row. We are told that among the 950 who passed the course, 96% of them passed the prelim, so the figure in cell (pass course, pass prelim) is (0.96)(950) = 912, and the number in cell (pass course fail prelim) = (0.04)(950) = 38. So, now our table looks like this:
$$\begin{array}{r|cc|r}
& \text{pass prelim}&\text{fail prelim} & \text{Total} \\ \hline
\text{pass course} & 912 &38 & 950 \\
\text{fail course} &? & ?& 50\\ \hline
\text{Total} & ? & ? & 1000
\end{array}
$$

By similar considerations you can fill in the rest of the table.

Now when you are asked for P(pass course|fail prelim) you are being asked asked for a ratio in the "fail prelim" column. All that is taken care of quickly and easily by Bayes' Theorem, but you can do the problem without it---it just takes a bit more work.

 

[FactChecker]

Homework Statement

View attachment 215217

Homework Equations

The Attempt at a Solution

View attachment 215218 [/B]
View attachment 215218

The % of course passing students who fail the prelim test is 38%. So, the required probability is 38 %.
Isn't the probability of a student failing the prelim when he his passing the course is known already = probability of a student who failed the prelim but pass the course?

No. If I am interpreting your phrasing correctly, they are not the same. To use more standard phrasing: The probability of a student failing the prelim given he passed the course ≠ probability of a student failing the prelim and passing the course. The first phrase looks at probabilities only within the group that has failed the prelim. The second phrase looks at probabilities within the entire space of students.

Another idea is to calculate total no. of students who failed the prelim let's say, N(pre F).
Then the required probability is $\frac {\text{no. of students who have passed the course but failed the prelim i.e. N(course P, prelim F)} }{\text{ non. of students who failed the prelim i.e.N( pre F)}$

Let's take the total no. of students to be 100.N( pre F) = 100 *95%*4% +100 * 5% * 75%= 7.55
N(course P, pre F) = 100 *95%*4%=3.8
Thus, the req. probability = 3.8/7.55=0.503

I have solved the question, but I don't understand what I had done here.

This is a fairly standard application of Bayes' Rule.

 

[FactChecker]

Bayes' Rule is important to know. It tells you how to adjust probabilities when you get some additional information. It solves a lot of problems methodically where your intuition can be very misleading.

 

[Pushoam]

Thank you.
I learned the problem.