1
$\begingroup$

I tried to solve this math problem in two different ways but found two different answers and I don't know which one is the right answer.

The problem goes like this:

The probability of failing in Science exam for a student is $\frac{1}{5}$, probability of passing in both Science and English is $\frac{3}{4}$ and probability of passing any one of these two subjects or both is $\frac{7}{8}$, what is the probability of him passing only in English?

$\mathbf{Solution:}$ Let S and E are the events of passing in Science and English respectively. probability of failing in Science $P(S^c) = \frac{1}{5}$, so probability of passing in Science exam is, $$P(S) = 1-P(S^c)=1-\frac{1}{5}=\frac{4}{5}$$ probability of passing both subjects, $P(S\cap E)= \frac{3}{4}$;

and probability of passing one or both subjects, $P(S\cup E)=\frac{7}{8}$

Since these two events are independent and not mutually exclusive, we can write, $$P(S\cup E)=P(S)+P(E)-P(S\cap E)$$ Where P(E) is the probability of passing in English. Which is $$P(E)=P(S\cup E)-P(S)+P(S\cap E)$$ $$P(E)=\frac{7}{8}-\frac{4}{5}+\frac{3}{4}$$ $$P(E) = \frac{33}{40}$$ At this point I can calculate the probability of him passing only in English in two ways:

(1) The probability of him passing only in English is $P(E\cap S^c)$

$$P(E)=P(E\cap S)+P(E\cap S^c)$$ Addition rule, since these two events are mutually exclusive. So, $$P(E\cap S^c)=P(E)-P(S\cap E)=\frac{33}{40}-\frac{3}{4}=\frac{3}{40}$$

(2)The probability of him passing only in English is $P(E\cap S^c)$.

Since these two events are independent, $$P(E\cap S^c)=P(E)P(S^c)=\frac{33}{40}\times\frac{1}{5}=\frac{33}{200}$$

This two methods give two different answers $\frac{3}{40}$ and $\frac{33}{200}$

I am obviously missing something or I have some lack in understanding. It will be very helpful if someone point out which one is wrong and why.

$\endgroup$
2
$\begingroup$

The key point is that $E$ and $S$ are dependent. Thus $$P(E\cap S^c)=P(E)\cdot P(S^c|E)=\frac{33}{40}\cdot \frac{1}{11}=\frac{3}{40}$$

You can comprehend $P(S^c|E)$ by looking at the table below:


\begin{array}{c|c|c|c} &E & E^c& \\ \hline S &\color{red}{0.75} & 0.05 & 0.8 \\ \hline S^c & 0.075 & \color{blue}{0.125} &\color{red}{0.2} \\ \hline & 0.825 & 0.175 &\color{red}{1} \end{array}

and probability of passing any one of these two subjects or both is $\frac78$

That means that $P(S^c\cap E^c)=1-\frac78=\frac18=\color{blue}{0.125}$

Finding out the missing figures is just simplest algebra.


You see that $P(S^c)=0.2 \neq P(S^c|E)$. Thus $E$ and $S$ are not independent. And your first approach is right $\checkmark$

$\endgroup$
  • $\begingroup$ Thank you for your help. But would you like to show me in detail how to get the table? Or is there any easy way to figure out the dependency of events? I thought events like in the problem is independent by common sense and that's how I generally used to approach any problem. Thanks in advance. $\endgroup$ – Arafat Hossen Oct 2 '18 at 3:33
  • $\begingroup$ In general if I have only two events I firstly make a table, almost always. The red figures are given in the text. And then the missing figures can be filled in successively. Especially the 0.8 should be very obvious since the cells of the last column/row sum up to one. I´ve made an edit. Please have a look. $\endgroup$ – callculus Oct 2 '18 at 5:03
  • $\begingroup$ @ArafatHossen You always have to prove if events are independent. There are two variants of the criteria. Basically the meaning is identical: 1. $P(A\cap B)=P(A)\cdot P(B)$, 2. $P(A|B)=P(A)$. It works as well by using the complements: $P(A\cap \overline B)=P(A)\cdot P(\overline B)\Rightarrow $ The events A and B are independent. $\endgroup$ – callculus Oct 2 '18 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.