If two events A and B are such that P(\(\bar A\)) = 0.3.P(B) = 0.4 and

1.	If two events A and B are such that P(\(\bar A\)) = 0.3.P(B) = 0.4 and P(A ∩ \(\bar B\)) = 0.5, find \(P(\cfrac{B}{\bar A \cap\bar B})\)
Answer» Given P(\(\bar A\)) = 0.3.P(B) = 0.4 and P(A ∩ \(\bar B\)) = 0.5 We know that P(A ∩ \(\bar B\)) = 0.7 - 0.5 = 0.2 Now, P(A∪B)=P(A) + P(B) - P(A∩B) = 0.7 + 0.4 - 0.4 = 0.9 Therefore, \(P(\overline{A\cup B})\) = 1 - P(A ∪ B) =1 - 0.9 = 0.1 ⇒ \(P(\cfrac{B}{\bar A \cap\bar B})\) = \(\cfrac14\)

If two events A and B are such that P(\(\bar A\)) = 0.3.P(B) = 0.4 and P(A ∩ \(\bar B\)) = 0.5, find \(P(\cfrac{B}{\bar A \cap\bar B})\)

Answer»

Given P(\(\bar A\)) = 0.3.P(B) = 0.4 and P(A ∩ \(\bar B\)) = 0.5

We know that P(A ∩ \(\bar B\)) = 0.7 - 0.5 = 0.2

Now, P(A∪B)=P(A) + P(B) - P(A∩B)

= 0.7 + 0.4 - 0.4

= 0.9

Therefore,

\(P(\overline{A\cup B})\) = 1 - P(A ∪ B)

=1 - 0.9

= 0.1

⇒ \(P(\cfrac{B}{\bar A \cap\bar B})\) = \(\cfrac14\)

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If two events A and B are such that P(\(\bar A\)) = 0.3.P(B) = 0.4 and P(A ∩ \(\bar B\)) = 0.5, find \(P(\cfrac{B}{\bar A \cap\bar B})\)

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