A) 0.4
B) 0.3
C) 0.2
D) 0.6
Correct Answer: D
Solution :
Let \[A\] denotes the event that the student is selected in IIT entrance test and \[B\] denotes the event that he is selected in Roorkee entrance test. Then \[P(A)=0.2,\,\,P(B)=0.5\] and \[P(A\cap B)=0.3.\] Required probability \[=P(\bar{A}\cap \bar{B})=1-P(A\cup B)\] \[=1-(P(A)+P(B)-P(A\cap B))\]\[=1-(0.2+0.5-0.3)=0.6\].You need to login to perform this action.
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