Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 34

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

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question