# Math Help - probability of success

1. ## probability of success

I am trying to answer this problem for an exam.Can anyone please help me.I wonder if I should be using the formula prob=nCr (p^r) q^(n-r).

The probability for students from a certain university to pass Mathematics and Economics are 3/7 and 5/7 respectively. If one student fails both subjects and there are four students who passed both subjects find the number of students who took the test.

2. Hello, hven191!

This doesn't require any fancy formulas . . .

The probability for students at a certain university
to pass Mathematics and Economics are 3/7 and 5/7 respectively.

One student failed both subjects and four students passed both subjects.

Find the number of students who took the test.
Make a Venn diagram of the students who passed.
Code:
      *---------------------------------------*
|                                       |
|   *-----------------------*           |
|   | Math                  |           |
|   | only  *---------------+-------*   |
|   |   M   |     Both      |       |   |
|   |       |       4       |       |   |
|   |       |               |       |   |
|   *-------+---------------*  Eco  |   |
|           |             E   only  |   |
|           *-----------------------*   |
|  Neither 1                            |
*---------------------------------------*

Let $M$ = number of students that passed Math only.
Let $E$ = number of students that passed Economics only.

We know there are 4 that passed both, and 1 that failed both.

The total number of students is: . $N \:=\:M + E + 5$

The number of students that passed Math is: . $M + 4$
The probability that a student passed Math is $\tfrac{3}{7}$
We have: . $\frac{M+4}{M+E+5} \:=\:\frac{3}{7} \quad\Rightarrow\quad 4M - 3E \:=\:-13\;\;{\color{blue}[1]}$

The number of students that passed Economics is $E + 4$
The probability that a student passes Economics is $\tfrac{5}{7}$
We have: . $\frac{E+4}{M+E+5} \:=\:\frac{5}{7} \quad\Rightarrow\quad 5M - 2E \:=\:3\;\;{\color{blue}[2]}$

$\begin{array}{cccccc}\text{Multiply {\color{blue}[1]} by -2:} & \text{-}8M + 6E &=& 26 & {\color{blue}[3]}\\
\text{Multiply {\color{blue}[2]} by 3:} & 15M - 6E &=& 9 & {\color{blue}[4]}\end{array}$

$\text{Add {\color{blue}[3]} and {\color{blue}[4]}: }\;7M \:=\:35 \quad\Rightarrow\quad\boxed{ M \:=\:5}$

Substitute into [1]: . $4(5) - 3E \:=\:-13 \quad\Rightarrow\quad\boxed{ E \:=\:11}$

Therefore, the number of students is: . $N \;=\;M + E + 5 \;=\;5 + 11 + 5 \;=\;{\color{red}21}$

3. ## Thanks

Thanks a lot Soroban! The solution you gave was very simple and easy to understand.The illustration was a big help. Although the answer does not match the answer in the book I'm using I'm still convinced with your answer. The answer in the book which is 28 may just be a typo-graphical error. Thanks!