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  4. If P(A)=(1/12), P(B)=(5/12), and P(B|A) =(1/15) then p(A∪ B) is equal to

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Q. If $ P\left(A\right)=\frac{1}{12}, P\left(B\right)=\frac{5}{12},$ and $P\left(B|A\right) =\frac{1}{15}$ then $p\left(A\cup B\right)$ is equal to

VITEEEVITEEE 2007Probability - Part 2

Solution:

Given
$P\left(A\right)=\frac{1}{12},P\left(B\right)=\frac{5}{12}, P\left(B/A\right)=\frac{1}{15}$
We know that $P\left(B/A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$
$\Rightarrow \frac{1}{15}=\frac{P\left(A\cap B\right)}{1/12}$
$\Rightarrow P\left(A\cap B\right)=\frac{1}{15\times12}=\frac{1}{180}$
But,
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
$\Rightarrow P\left(A\cup B\right)=\frac{1}{12}+\frac{5}{12}-\frac{1}{180}=\frac{89}{180}$