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(C) Let $P(A) = 1/5$ be the probability of $A$ failing and $P(B) = 3/10$ be the probability of $B$ failing.Assuming $A$ and $B$ are independent events

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$19/50$

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(C) Let $P(A) = 1/5$ be the probability of $A$ failing and $P(B) = 3/10$ be the probability of $B$ failing.Assuming $A$ and $B$ are independent events,the probability that either $A$ or $B$ fails (but not both) is given by $P(A \cup B) - P(A \cap B)$.Alternatively,the probability that exactly one of them fails is $P(A \cap \overline{B}) + P(\overline{A} \cap B)$.$P(A \cap \overline{B}) = P(A) 	imes (1 - P(B)) = \frac{1}{5} 	imes (1 - \frac{3}{10}) = \frac{1}{5} 	imes \frac{7}{10} = \frac{7}{50}$.$P(\overline{A} \cap B) = (1 - P(A)) 	imes P(B) = (1 - \frac{1}{5}) 	imes \frac{3}{10} = \frac{4}{5} 	imes \frac{3}{10} = \frac{12}{50}$.Total probability $= \frac{7}{50} + \frac{12}{50} = \frac{19}{50}$.

If the probability of $A$ failing an exam is $1/5$ and the probability of $B$ failing is $3/10$,what is the probability that either $A$ or $B$ fails?

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