A) \[\frac{41}{100}\]
B) \[\frac{33}{100}\]
C) \[\frac{1}{10}\]
D) None of these
Correct Answer: B
Solution :
Let \[A\] be the event to be multiple of 4 and \[B\] be the event to be multiple of 6 So, \[P(A)=\frac{25}{100},\] \[P(B)=\frac{16}{100}\] and \[P(A\cap B)=\frac{8}{100}\] Thus required probability is \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow P(A\cup B)=\frac{25}{100}+\frac{16}{100}-\frac{8}{100}=\frac{33}{100}\].You need to login to perform this action.
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