Three Mutually Exclusive Events

[nasi112]

Three mutually exclusive events occur with probabilities [MATH]P(E_1) = 0.25, P(E_2) = 0.35, [/math] and [math]P(E_3) = 0.40[/MATH]. Other probabilities are [MATH]P(B|E_1) = 0.24, P(B|E_2) = 0.18, P(B|E_3) = 0.58[/MATH]
Find [MATH]P(E_1|B)[/MATH].

[MATH]P(B|E_1) = \frac{P(B \cap E_1)}{P(E_1)}[/MATH]
[MATH]P(B \cap E_1) = P(B|E_1)P(E_1) = P(E_1 \cap B)[/MATH]
now

[MATH]P(E_1|B) = \frac{P(E_1 \cap B)}{P(B)} = \frac{P(B|E_1)P(E_1)}{P(B)}[/MATH]
I am stuck here because I don't have the probability, [MATH]P(B)[/MATH]

 

[laganvishu]

Directly use the formula given by Bayes'
$\displaystyle P(E_1|B)=\frac{P(B|E_1)P(E_1)} {P(B|E_1)P(E_1)+P(B|E_2)P(E_2)+P(B|E_3)P(E_3)}$

 

[nasi112]

Thanks a lot. How did you derive [MATH]P(B)[/MATH]?

 

[Dr.Peterson]

Thanks a lot. How did you derive [MATH]P(B)[/MATH]?

The events [MATH]B \cap E_1, B \cap E_2, B \cap E_3[/MATH] are mutually exclusive and together constitute event B. The denominator [MATH]P(B|E_1)P(E_1)+P(B|E_2)P(E_2)+P(B|E_3)P(E_3)[/MATH] is the probability of their union, which is B.

 

[nasi112]

Thanks a lot Dr.Peterson. I started to grasp the idea.