In a class,30 % of students passed English,20 | vedclass

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(A) Let $E$ be the event that a student passed English and $F$ be the event that a student passed French.Given:$P(E) = \frac{30}{100} = \frac{3}{10}$$

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(A) Let $E$ be the event that a student passed English and $F$ be the event that a student passed French.Given:$P(E) = \frac{30}{100} = \frac{3}{10}$$P(F) = \frac{20}{100} = \frac{2}{10} = \frac{1}{5}$$P(E \cap F) = \frac{10}{100} = \frac{1}{10}$We need to find the probability that the student passed English or French,which is $P(E \cup F)$.Using the addition theorem of probability:$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$P(E \cup F) = \frac{3}{10} + \frac{2}{10} - \frac{1}{10}$$P(E \cup F) = \frac{3 + 2 - 1}{10} = \frac{4}{10} = \frac{2}{5}$

In a class,$30 \%$ of students passed English,$20 \%$ passed French and $10 \%$ passed both. If a student is selected at random,what is the probability that he has passed English or French?

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