The probabilities that a student passes in Ma | vedclass

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Question

(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}} = \

Vedclass · Accepted Answer

$p + m + c = \frac{{27}}{{20}}$

Vedclass · Answer

(b) Let $M,\,\,P$ and $C$ be the events of passing in mathematics, physics and chemistry respectively.

$P(M \cup P \cup C) = \frac{{75}}{{100}} = \frac{3}{4}$

$P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{{50}}{{100}} = \frac{1}{2}$

$P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{{40}}{{100}} = \frac{2}{5}$

$	herefore$ $m(1 - p)(1 - c) + p(1 - m)(1 - c) + c(1 - m)(1 - p)$

$ + 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 $(ii),$ $m + p + c - mpc = \frac{3}{4} + \frac{1}{2}$

$	herefore \,\,\,m + p + c = \frac{3}{4} + \frac{1}{2} + \frac{1}{{10}} = \frac{{15 + 10 + 2}}{{20}} = \frac{{27}}{{20}}.$

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