The events A and B are independent among three events A, B and D. If \(P(A\cap B\cap D) = 0.04\), \(P(D\mid A\cap B) = 0.25\) and \(P(B) = 4P(A)\), then what is the value of \(P(A\cup B)\)?
0.84
Since \(P(D\mid A\cap B)=\dfrac{P(A\cap B\cap D)}{P(A\cap B)}\), we get \(P(A\cap B) = \dfrac{0.04}{0.25} = 0.16\). As A and B are independent, \(P(A)P(B) = 0.16\), and with \(P(B)=4P(A)\) this gives \(4P(A)^2 = 0.16\), so \(P(A)=0.2\) and \(P(B)=0.8\). Hence \(P(A\cup B) = P(A)+P(B)-P(A\cap B) = 0.2+0.8-0.16 = 0.84\).
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