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An observed event $B$ can occur after one of the three events $A_{1}, A_{2}, A_{3} .$ If
$P\left(A_{1}\right)=P\left(A_{2}\right)=0.4, P\left(A_{3}\right)=0.2$ and $P\left(B / A_{1}\right)=0.25, P\left(B / A_{2}\right)$
$=0.4, P\left(B / A_{3}\right)=0.125$, what is the probability of $A_{1}$ after observing $B$ ?
Options:
$P\left(A_{1}\right)=P\left(A_{2}\right)=0.4, P\left(A_{3}\right)=0.2$ and $P\left(B / A_{1}\right)=0.25, P\left(B / A_{2}\right)$
$=0.4, P\left(B / A_{3}\right)=0.125$, what is the probability of $A_{1}$ after observing $B$ ?
Solution:
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Verified Answer
The correct answer is:
$\frac{20}{57}$
By Baye's theorem Required probability $=P\left(A_{1} / B\right)$ $=\frac{P\left(A_{1}\right) P\left(B / A_{1}\right)}{P\left(A_{1}\right) P\left(B / A_{1}\right)+P\left(A_{2}\right) P\left(B / A_{2}\right)+P\left(A_{3}\right) P\left(B / A_{3}\right)}$
$=\frac{0.4 \times 0.25}{0.4 \times 0.25+0.4 \times 0.4+0.2 \times 0.125}$
$=\frac{0.1}{0.1+0.16+0.025}=\frac{0.1}{0.285}=\frac{20}{57}$
$=\frac{0.4 \times 0.25}{0.4 \times 0.25+0.4 \times 0.4+0.2 \times 0.125}$
$=\frac{0.1}{0.1+0.16+0.025}=\frac{0.1}{0.285}=\frac{20}{57}$
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