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Classical Definition of Probability

Bag I contain...

Question

Bag $I$ contains 3 black and 2 white balls, bag $I I$ contains 2 black and 4 white balls, bag $I I I$ contains 4 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of getting a black ball.

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Solution

Let $E_{1}$ = Event that bag $I$ is selected
$E_{2}$ = Event that bag $I I$ is selected
$E_{3}$ = Event that bag $I I I$ is selected
$E$ = Event that selected ball is black
$P (E | E_{1}) = P (s e l e c t e d b a l l i s b l a c k g i v e n t h a t b a g I i s s e l e c t e d) = \frac{3}{5}$
$P (E | E_{2}) = P (s e l e c t e d b a l l i s b l a c k g i v e n t h a t b a g I I i s s e l e c t e d) = \frac{2}{6}$
$P (E | E_{3}) = P (s e l e c t e d b a l l i s b l a c k g i v e n t h a t b a g I I I i s s e l e c t e d) = \frac{4}{8}$
$P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$
$P (E) = P (E | E_{1}) \times P (E_{1}) + P (E | E_{2}) \times P (E_{2}) + P (E | E_{3}) \times P (E_{3})$
$P (E) = \frac{3}{5} \times \frac{1}{3} + \frac{2}{6} \times \frac{1}{3} + \frac{4}{8} \times \frac{1}{3} = \frac{43}{90}$

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