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.

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions