Question

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?

Answer 1

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

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

And H = The event that selected person is left-handed

∴ P(E₁) = 0.30 and P(E₂) = 0.70

`"P"("H"/"E"_1)` = 0.06

`"P"("H"/"E"_2)` = 0.10

So, from Bayes’ Theorem

`"P"("E"_1/"H") = ("P"("E"_1)*"P"("H"/"E"_1))/("P"("E"_1)*"P"("H"/"E"_1) + "P"("E"_2)*"P"("H"/"E"_2))`

= `(0.30 xx 0.06)/(0.30 xx 0.06 + 0.70 xx 0.10)`

= `(0.018)/(0.018 + 0.070)`

= `0.018/0.088`

= `9/44`

Hence, the required probability is `9/44`.