91Ó°ÊÓ

Suppose that 80 percent of all statisticians are shy, whereas only 15 percent of all economists are shy. Suppose also that 90 percent of the people at a large gathering are economists and the other 10 percent are statisticians.

Let define events,

A: Gathering are Statistician

B: Gathering are Economics are shy

S: The person is shy

\(\begin{aligned}{}\Pr \left( A \right) &= 0.10\\\Pr \left( B \right) &= 0.90\\\Pr \left( {S|A} \right) &= 0.80\\\Pr \left( {S|B} \right) &= 0.15\end{aligned}\)

If one meet a shy person at random at the gathering, what is the probability that the person is a statistician obtained as:

\(\begin{aligned}{}i.e,\Pr \left( {A|S} \right) &= \frac{{\Pr \left( {A \cap S} \right)}}{{\Pr \left( S \right)}}\\ &= \frac{{\Pr \left( A \right)\Pr \left( {S|A} \right)}}{{\Pr \left( A \right)\Pr \left( {S|A} \right) + \Pr \left( B \right)\Pr \left( {S|B} \right)}}\\ &= \frac{{0.10 \times 0.80}}{{0.10 \times 0.80 + 0.90 \times 0.15}}\\ &= 0.372093\end{aligned}\)

The probability that the person is a statistician is 0.372093.