If P(B) = \(\frac{3}{5} \)and P(A’ ∩ B) = \(\frac{1}{2}\) for

1.	If P(B) = \(\frac{3}{5} \)and P(A’ ∩ B) = \(\frac{1}{2}\) for two events A and B, find P(A\|B).
Answer» P(B) = \(\frac{3}{5}\)and P(A’ ∩ B) = \(\frac{1}{2}\) are given. ∴ P(A’ ∩ B) = P(B) – P(A ∩ B) ∴ \(\frac{1}{2} = \frac{3}{5}\) – P(A ∩ B) ∴ P(A ∩ B) = \(\frac{3}{5} – \frac{1}{2} = \frac{6−5}{10 }= \frac{1}{10}\) Now, P(A\|B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{1}{10}\frac{3}{5} = \frac{1}{10}×\frac{5}{3} = \frac{1}{6}\) ∴ P(A\|B) = \(\frac{1}{6}\)

If P(B) = \(\frac{3}{5} \)and P(A’ ∩ B) = \(\frac{1}{2}\) for two events A and B, find P(A|B).

Answer»

P(B) = \(\frac{3}{5}\)and P(A’ ∩ B) = \(\frac{1}{2}\) are given.

∴ P(A’ ∩ B) = P(B) – P(A ∩ B)

∴ \(\frac{1}{2} = \frac{3}{5}\) – P(A ∩ B)

∴ P(A ∩ B) = \(\frac{3}{5} – \frac{1}{2} = \frac{6−5}{10 }= \frac{1}{10}\)

Now, P(A|B) = \(\frac{P(A∩B)}{P(B)}\)

= \(\frac{1}{10}\frac{3}{5} = \frac{1}{10}×\frac{5}{3} = \frac{1}{6}\)

∴ P(A|B) = \(\frac{1}{6}\)

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If P(B) = \(\frac{3}{5} \)and P(A’ ∩ B) = \(\frac{1}{2}\) for two events A and B, find P(A|B).

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