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  • 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 bl

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?

 

Answers (1)

Given-

\begin{array}{|l|l|l|} \hline & \begin{array}{l} \text { Blood group } \\ \text { 'O } \end{array} & \begin{array}{l} \text { Other than } \\ \text { blood group } \\ \text { 'O' } \end{array} \\ \hline \begin{array}{l} \text { I. Number of } \\ \text { people } \end{array} & 30 \% & 70 \% \\ \hline \begin{array}{l} \text { II. Percentage } \\ \text { of left-handed } \\ \text { people } \end{array} & 6 \% & 10 \% \\ \hline \end{array}

Let E1 be the event that the person selected is of group O
E2 be the event that the person selected is of other than blood group O
And E3 be the event that the person selected is left handed
∴P(E1) =0.30, P(E2) =0.70
P(E3|E1) = 0.060 And P(E3|E2) =0.10
Using bayes’ theorem, we have:

\\ P\left(E_{1} \mid E_{3}\right)=\frac{P\left(E_{1}\right) \cdot P\left(E_{3} \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(E_{3} \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E_{3} \mid E_{2}\right)} \\ =\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}

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