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The probabilities that a student passes in Mathematics, Physics and Chemistry are $m, p$ and $c$ respectively. On these subjects, the student has a $75\%$ chance of passing in at least one, a $50\%$ chance of passing in at least two and a $40\%$ chance of passing in exactly two. Which of the following relations are true
$p + m + c = \frac{{19}}{{20}}$
$p + m + c = \frac{{27}}{{20}}$
$pmc = \frac{1}{{10}}$
$pmc = \frac{1}{4}$
Solution
(b) Let $M,\,\,P$ and $C$ be the events of passing in mathematics, physics and chemistry respectively.
$P(M \cup P \cup C) = \frac{{75}}{{100}} = \frac{3}{4}$
$P(M \cap P) + P(P \cap C) + P(M \cap C) – 2P(M \cap P \cap C) = \frac{{50}}{{100}} = \frac{1}{2}$
$P(M \cap P) + P(P \cap C) + P(M \cap C) – 2P(M \cap P \cap C) = \frac{{40}}{{100}} = \frac{2}{5}$
$\therefore$ $m(1 – p)(1 – c) + p(1 – m)(1 – c) + c(1 – m)(1 – p)$
$ + mp(1 – c) + mc(1 – p) + pc(1 – m) + mpc = \frac{3}{4}$
$ \Rightarrow m + p + c – mc – mp – pc + mpc = \frac{3}{4}$ …..$(i)$
Similarly, $mp(1 – c) + pc(1 – m) + mc(1 – p) + mpc = \frac{1}{2}$
$ \Rightarrow mp + pc + mc – 2mpc = \frac{1}{2}$ …..$(ii)$
$mp(1 – c) + pc(1 – m) + mc(1 – p) = \frac{2}{5}$
$ \Rightarrow mp + pc + mc – 3mpc = \frac{2}{5}$ …..$(iii)$
From $(ii)$ to $(iii),$ $mpc = \frac{1}{2} – \frac{2}{5} = \frac{1}{{10}}$
From $(i)$ and $(ii),$ $m + p + c – mpc = \frac{3}{4} + \frac{1}{2}$
$\therefore \,\,\,m + p + c = \frac{3}{4} + \frac{1}{2} + \frac{1}{{10}} = \frac{{15 + 10 + 2}}{{20}} = \frac{{27}}{{20}}.$
