Suppose that 6% of the people with blood group O are left-handed and 10%

Let’s assume

E1 = The event that a person selected is of blood group O

E2 = The event that the people selected is of other group

And H = The event that selected person is left handed

Now,

P(E1) = 0.30 and P(E2) = 0.70

Further, P(H/E1) = 0.06 and P(H/E2) = 0.10

Using Baye’s Theorem, we have

Therefore, the required probability is 9/44.