The probability that a student will succeed i | vedclass

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(D) Let $A$ be the event that the student is selected in the $IIT$ entrance test and $B$ be the event that the student is selected in the Roorkee entr

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(D) Let $A$ be the event that the student is selected in the $IIT$ entrance test and $B$ be the event that the student is selected in the Roorkee entrance test.Given: $P(A) = 0.2$,$P(B) = 0.5$,and $P(A \cap B) = 0.3$.We need to find the probability that he does not succeed at either place,which is $P(\overline{A} \cap \overline{B})$.By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.Using the Addition Theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = 0.2 + 0.5 - 0.3 = 0.4$.Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.4 = 0.6$.

The probability that a student will succeed in the $IIT$ entrance test is $0.2$ and the probability that he will succeed in the Roorkee entrance test is $0.5$. If the probability that he will be successful at both places is $0.3$,then the probability that he does not succeed at either place is:

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