Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 33 maths textbook solution

Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 33 maths textbook solution

Answers (1)

Answer:

 \frac{90}{589}

Hint:

 Use Baye’s theorem.

Given:

 The test will correctly detect the disease 1% of time probability large population of which an estimate 0.2% has disease a person.

Solution:

Let A,E1 and E2  denote the event that the person suffer from the disease, the test detecte the disease correctly and the test does not detect the disease correctly, respectively.

\begin{aligned} &P(E_1)=0.002\\ &P(E_2)=0.998\\ \end{aligned}

Now

\begin{aligned} &P\left ( \frac{A}{E_1} \right )=0.90\\ &P\left ( \frac{A}{E_2} \right )=0.01\\ \end{aligned}

Using Baye’s theorem we get

\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 )}\\ &=\frac{{\frac{90}{100}\times \frac{2}{1000} }}{\frac{90}{100}\times \frac{2}{1000}+\frac{1}{100}\times \frac{998}{1000}}\\ &=\frac{180}{180+998}\\ &=\frac{180}{1178}\\ &=\frac{90}{589} \end{aligned}

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