Under throwing a die one time, if the number 3 does not appear what is the probability of getting odd numbers?
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total answers (1)
total answers (1)
Statistics
The sample space is S={1,2,3,4,5,6}
need an explanation for this answer? contact us directly to get an explanation for this answerLet A be the event that number 3 does not appear
A={1,2,4,5,6}
And B is the event that odd numbers appear
B={1,3,5} ,then:
P(B│A)=(P(A∩B))/(P(A))
We have
A∩B={1,5}→P(A∩B)=2/6
P(A)=5/6
P(B│A)=(2/6)/(5/6)=2/5