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.
Given:
$P(M \cup P \cup C) = \frac{75}{100} = \frac

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.
Given:
$P(M \cup P \cup C) = \frac{75}{100} = \frac{3}{4}$
$P(	ext{at least two}) = P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{50}{100} = \frac{1}{2}$
$P(	ext{exactly two}) = P(M \cap P) + P(P \cap C) + P(M \cap C) - 3P(M \cap P \cap C) = \frac{40}{100} = \frac{2}{5}$
Subtracting the two equations:
$P(M \cap P \cap C) = \frac{1}{2} - \frac{2}{5} = \frac{1}{10}$
Using the inclusion-exclusion principle:
$P(M \cup P \cup C) = (m + p + c) - (mp + pc + mc) + mpc = \frac{3}{4}$
From the 'at least two' condition:
$(mp + pc + mc) - 2mpc = \frac{1}{2} \Rightarrow (mp + pc + mc) = \frac{1}{2} + 2(\frac{1}{10}) = \frac{7}{10}$
Substituting these into the inclusion-exclusion equation:
$(m + p + c) - \frac{7}{10} + \frac{1}{10} = \frac{3}{4}$
$m + p + c = \frac{3}{4} + \frac{6}{10} = \frac{15 + 12}{20} = \frac{27}{20}$.

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