Suppose that 6% of the people with blood group $O$ are left handed and 10% of those with other blood groups are left handed, 30% of the people have blood group $O$. If a left handed person is selected at random, what is the probability that he/she will have blood group $O$?

$\frac{9}{44}$

The correct answer is Option (1) → $\frac{9}{44}$ ##

Let $E_1 = \text{Event that the person selected is of blood group } O$

$ E_2 = \text{Event that the person selected is of other than blood group } O$

and $E_3 = \text{Event that selected person is left handed.}$

$∴P(E_1) = 0.30, P(E_2) = 0.70$

$P(E_3 | E_1) = 0.06 \text{ and } P(E_3 | E_2) = 0.10$

By using Baye's theorem, $P(E_1 / E_3) = \frac{P(E_1) \cdot P(E_3 | E_1)}{P(E_1) \cdot P(E_3 | E_1) + P(E_2) \cdot P(E_3 | E_2)}$

$= \frac{0.30 \times 0.06}{(0.30 \times 0.06) + (0.70 \times 0.10)}$

$= \frac{0.0180}{0.0180 + 0.0700} = \frac{0.0180}{0.0880} = \frac{180}{880} = \frac{9}{44}$