The probabilities that a student passes in Ma | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:
$(m + p + c) - \frac{7}{10} + \frac{1}{10} = \frac{3}{4}$
$m + p + c = \frac{3}{4} + \frac{6}{10} = \frac{15 + 12}{20} = \frac{27}{20}$.

Similar Questions

If $X$ and $Y$ are two sets such that $n(X) = 17$,$n(Y) = 23$ and $n(X \cup Y) = 38$,find $n(X \cap Y)$.

In a certain test,$a_i$ students gave wrong answers to at least $i$ questions,where $i = 1, 2, 3, \dots, k$. No student gave more than $k$ wrong answers. The total number of wrong answers given is:

There are $200$ individuals with a skin disorder. $120$ had been exposed to the chemical $C_{1}$,$50$ to chemical $C_{2}$,and $30$ to both the chemicals $C_{1}$ and $C_{2}$. Find the number of individuals exposed to chemical $C_{1}$ or chemical $C_{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module