Given, P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.3,then \(P\big(\frac{A'}{B

1.	Given, P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.3,then \(P\big(\frac{A'}{B'}\big)\) is equal to(a) \(\frac{1}{3}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{3}{4}\)
Answer» Answer: (C) \(\frac{2}{3}\) \( P\big(\frac{A'}{B'}\big) = \frac{P(A'\,\cap\,B')}{P(B')} = \frac{P(A\,\cap\,B)'}{P(B')}\) = \(\frac{1-P(A\,\cup\,B)}{1-P(B)} = \frac {1-[P(A)+P(B)-P(A\,\cap\,B)]}{1-P(B)}\) \(= \frac{1-[0.5+0.4-0.3]}{1-0.4}=\frac{0.4}{0.6}=\frac{2}{3}\)

Given, P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.3,then \(P\big(\frac{A'}{B'}\big)\) is equal to(a) \(\frac{1}{3}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{3}{4}\)

Answer»

Answer: (C) \(\frac{2}{3}\)

\( P\big(\frac{A'}{B'}\big) = \frac{P(A'\,\cap\,B')}{P(B')} = \frac{P(A\,\cap\,B)'}{P(B')}\)

= \(\frac{1-P(A\,\cup\,B)}{1-P(B)} = \frac {1-[P(A)+P(B)-P(A\,\cap\,B)]}{1-P(B)}\)

\(= \frac{1-[0.5+0.4-0.3]}{1-0.4}=\frac{0.4}{0.6}=\frac{2}{3}\)

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Given, P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.3,then \(P\big(\frac{A'}{B'}\big)\) is equal to(a) \(\frac{1}{3}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{3}{4}\)

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