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Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 34

Answers (1)

Answer:

 D1

Hint:

 Use baye’s theorem.

Given:

 1800 had disease d1 , 2100 has disease d2 , and the other disease d3.1500 patient with disease d1, 1200 disease d3 is 900.  

Solution:

Let  A, E1,E2,E3  denote the events that the patient show symptom patient has disease d1 , has disease d2  and has disease d3 , respectively.

\begin{aligned} &P(E_1)=\frac{1800}{5000}\\ &P(E_2)=\frac{2100}{5000}\\ &P(E_3)=\frac{1100}{5000}\\ \end{aligned}

Now,

\begin{aligned} &P\left ( \frac{A}{E_1} \right )=\frac{1500}{1800}\\ &P\left ( \frac{A}{E_2} \right )=\frac{1200}{2100}\\ &P\left ( \frac{A}{E_3} \right )=\frac{9000}{1100}\\ \end{aligned}

Using Baye’s theorem we get

Required probability

\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1800}{5000}\times \frac{1500}{1800} }}{\frac{1800}{5000}\times \frac{1500}{1800}+\frac{2100}{5000}\times \frac{1200}{2100}+\frac{1100}{5000}\times \frac{9000}{1100}}\\ &=\frac{15}{15+12+9}\\ &=\frac{15}{12} \end{aligned}

Required probability

\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{2100}{5000}\times \frac{1200}{2100} }}{\frac{1800}{5000}\times \frac{1500}{1800}+\frac{2100}{5000}\times \frac{1200}{2100}+\frac{1100}{5000}\times \frac{9000}{1100}}\\ &=\frac{12}{15+12+9}\\ &=\frac{12}{36} \end{aligned}

Required probability

\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1100}{5000}\times \frac{9000}{1100} }}{\frac{1800}{5000}\times \frac{1500}{1800}+\frac{2100}{5000}\times \frac{1200}{2100}+\frac{1100}{5000}\times \frac{9000}{1100}}\\ &=\frac{9}{15+12+9}\\ &=\frac{9}{36}\\ &=\frac{1}{4} \end{aligned}

As

\begin{aligned} &P\left (\frac{E_1}{A} \right ) \end{aligned}

is maximum, so, it is most likely that the person suffer from the disease d1.

Posted by

Gurleen Kaur

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