Question

The probabilities that a student passes in mathematics, physics and chemistry are $m,p$ and $c$ respectively. Of these subjects, the students have a $75\%$, chance of passing in atleast one, a $50\%$ chance of passing in at least two and a $40\%$ chance of passing in exactly two. Which of the following relation(s) is/are true?

• A. $p+m+c=\frac{19}{20}$
• B. $p+m+c=\frac{27}{20}$
• C. $pmc=\frac{1}{10}$
• D. $pmc=\frac{1}{4}$

Answer 1

Answer: (B), (C)

$pmc=\frac{1}{10}$

$P\left(\text{passing in atleast one subject}\right)=\frac{75}{100}\Rightarrow m+p+c-\sum mp+mpc=\frac{3}{4}\cdots \left(i\right)$
$P$(passing in atleast $2$ subjects) $=\sum mp-2mpc=\frac{50}{100}=\frac{1}{2}\cdots \left(ii\right)$
$P$(passing exactly $2$ subjects) $=\sum mp-3mpc=\frac{40}{100}=\frac{2}{5}\cdots \left(iii\right)$
From equation $\left(ii\right)$ and $\left(iii\right),$ we have
$mpc=\frac{1}{2}-\frac{2}{5}=\frac{1}{10}$
$\therefore$ From $\left(ii\right)$ $\sum mp=\frac{1}{2}+2mpc=\frac{1}{2}+\frac{1}{5}=\frac{7}{10}$
Now, from $\left(i\right)$
$m+p+c=\frac{3}{4}+\sum mp-mpc=\frac{3}{4}+\frac{7}{10}-\frac{1}{10}=\frac{15+14-2}{20}=\frac{27}{20}$