Question

Suppose there are two bowls full of biscuits. Bowl-I has 10 chocolate biscuits and 30 cream biscuits, while bowl-II has 20 of each. Seeta picks a bowl at random, and then picks a biscuit at random. We may assume there is no reason to believe Seeta treats one bowl differently from another, likewise for the biscuits. The biscuit turns out to be a cream one. What is the probability that Seeta picked it out of bowl I?

Answer 1

Event of choosing a bowl is B

So,probability of choosing any bowl is
$p(B) = \frac{1}{2}$
Let Probability of choosing cream biscuit from bowl 1 is
$p(C1) = \frac{30}{40} \\ \\ = \frac{3}{4}$
Let Probability of choosing cream biscuit from bowl 2 is
$p(C2) = \frac{20}{40} \\ \\ = \frac{1}{2}$

Now Probability of choosing a cream biscuit from bowl I is calculated by Bay's theorem

$= \frac{p(B)p(C1)}{p(B)p(C1) + p(B)p(C2)} \\ \\ = \frac{ \frac{1}{2} \times \frac{3}{4} }{ \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{1}{2} } \\ \\ = \frac{ \frac{3}{4} }{ \frac{3 + 2}{4} } \\ \\ = \frac{3}{5}$

Thus,probability that Seeta picked a cream biscuit out of bowl I is
$= \frac{3}{5}$

Hope it helps you