# 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?

Let$$E_1$$ be a person selected be 'o' group
$$E_2$$ be a person selected be of non 'o' group
A; He is left handed person
P($$E_1/A$$)=P($$he\; is\; 'o' \;group/he \;is\; left \;handed$$)
P(E$$_1$$/A)$$\Large\frac{{p(E_1)}{P(A/E_1)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}$$

P$$E_1$$=P(He is 'o' group)=$$\large\frac{30}{100}$$=$$\large\frac{3}{10}$$
P$$E_2$$=P(He isnot 'o' group)=$$\large\frac{70}{100}$$=$$\large\frac{7}{10}$$
P($$A/E_1$$)=P(He is left handed /he is'O' groyp)
=$$\large\frac{6}{100}$$
P($$A/E_2$$)=P(He is left handed /he is not 'O' groyp)
=$$\large\frac{10}{100}$$
P($$E_1/A$$)=$$\Large\frac{\frac{3}{10}\times\frac{6}{100}}{\frac{3}{10}\times\frac{6}{100}+\frac{7}{10}\times\frac{10}{100}}=\frac{3\times6}{3\times6+7\times10}$$
=$$\large\frac{18}{88}$$
=$$\large\frac{9}{44}$$

edited Jun 4, 2013