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(C) Let $P(A)$,$P(B)$,and $P(C)$ be the probabilities of passing in the first,second,and third grade respectively.Given $P(A) = 1/10$,$P(B) = 3/5$,and

Vedclass · Accepted Answer

$1/20$

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(C) Let $P(A)$,$P(B)$,and $P(C)$ be the probabilities of passing in the first,second,and third grade respectively.Given $P(A) = 1/10$,$P(B) = 3/5$,and $P(C) = 1/4$.Since these events are mutually exclusive,the probability of passing in any of these grades is $P(A \cup B \cup C) = P(A) + P(B) + P(C)$.$P(A \cup B \cup C) = 1/10 + 3/5 + 1/4 = (2 + 12 + 5) / 20 = 19/20$.The probability of failing (getting a fourth grade) is $P(D) = 1 - P(A \cup B \cup C)$.$P(D) = 1 - 19/20 = 1/20$.Converting to decimal,$1/20 = 0.05 = 5/100 = 1/20$.

If the probabilities of a student passing in the first,second,or third grade are $1/10$,$3/5$,and $1/4$ respectively,then the probability of the student failing (getting a fourth grade) is ........

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