In a college,40 % of students attend the Math | vedclass

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Question

(A) Let $M$ be the set of students attending the Mathematics class and $P$ be the set of students attending the Physics class. Given: $P(M) = 40 \%$,$

Vedclass · Accepted Answer

$\frac{3}{10}$

Vedclass · Answer

(A) Let $M$ be the set of students attending the Mathematics class and $P$ be the set of students attending the Physics class. Given: $P(M) = 40 \%$,$P(P) = 30 \%$,and $P(M \cap P) = 20 \%$. The probability of students attending only the Mathematics class is $P(M) - P(M \cap P) = 40 \% - 20 \% = 20 \%$. The probability of students attending only the Physics class is $P(P) - P(M \cap P) = 30 \% - 20 \% = 10 \%$. The probability that the student attends only one class is the sum of the probabilities of attending only Mathematics and only Physics: $20 \% + 10 \% = 30 \%$. Converting to a fraction,$30 \% = \frac{30}{100} = \frac{3}{10}$.

In a college,$40 \%$ of students attend the Mathematics class,$30 \%$ of students attend the Physics class,and $20 \%$ of students attend both classes. If a student is chosen at random from the college,what is the probability that the chosen student attends only one class?

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