Probability problem possibly based on principle of inclusion exclusion

Question

The problem reads as follows:

Probabilities that Rajesh passes in Physics, Math and Chemistry are $p$, $m$ and $c$ respectively. Of these subjects, Rajesh has $75%$ chance of passing in at least one, $50%$ chance of passing in at least two and $40%$ chance of passing in exactly two. Find which of the following is true:

(a) $p+m+c=\frac{19}{20}$
(b) $p+m+c=\frac{27}{20}$
(c) $pmc = \frac{1}{20}$
(d) $pmc = \frac{1}{8}$

Given answer is (b)

What all I am able to guess, is following

Given:

1. At least one = $75\%$
2. At least two = $50\%$
3. Exactly two = $40\%$

From this

4. Exactly three = At least two - Exactly two $= 50\%-40\% = 10\%$
5. Exactly one = At least one - At least two $= 75\% - 50\% = 25\%$
6. None = $100\%$ - At least one $=100\%-75\%=25\%$

However I dont find how to proceed.

Answer 1

Draw a Venn diagram, you will find that

$p+m+c =$ P(exactly one) + 2P(exactly two) + 3P(exactly three)

= $\dfrac{25 + 2\cdot40 + 3\cdot10}{100} = \dfrac{135}{100} = \dfrac{27}{20}$

Answer 2

Hint: You can see the following, possibly from a Venn diagram.

$p+m+c=P(\text{exactly one passed})+2\cdot P(\text{exactly two passed})+3\cdot P(\text{all three passed})$