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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?
Solution
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`.
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