(b) माना $M,\,\,P$ और $C$ क्रमश: गणित, भौतिक तथा रसायन में उत्तीर्ण होने की घटनायें हों तब

$P(M \cup P \cup C) = \frac{{75}}{{100}} = \frac{3}{4}$

$P(M \cap P) + P(P \cap C) + P(M \cap C) – 2P(M \cap P \cap C) = \frac{{50}}{{100}} = \frac{1}{2}$

$P(M \cap P) + P(P \cap C) + P(M \cap C) – 2P(M \cap P \cap C) = \frac{{40}}{{100}} = \frac{2}{5}$

$\therefore$ $m(1 – p)(1 – c) + p(1 – m)(1 – c) + c(1 – m)(1 – p)$

$ + mp(1 – c) + mc(1 – p) + pc(1 – m) + mpc = \frac{3}{4}$

$ \Rightarrow m + p + c – mc – mp – pc + mpc = \frac{3}{4}$ …..$(i)$

इसी प्रकार $mp(1 – c) + pc(1 – m) + mc(1 – p) + mpc = \frac{1}{2}$

$ \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}$ …..$(iii)$

$(ii)$ और $(iii)$ से, $mpc = \frac{1}{2} – \frac{2}{5} = \frac{1}{{10}}$

$(i)$ और $(ii)$ से, $m + p + c – mpc = \frac{3}{4} + \frac{1}{2}$

$\therefore \,\,\,m + p + c = \frac{3}{4} + \frac{1}{2} + \frac{1}{{10}} = \frac{{15 + 10 + 2}}{{20}} = \frac{{27}}{{20}}.$