Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 3

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 3

Answers (1)

Answer:

 \frac{29}{45}

Hint:

To solve this problem, we use 

\text { total probability }=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )

Given:

 One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls.

Solution:

E1 = red bag from 1 to 2

E2 = yellow bag from 1 to 2

A = ball is yellow

\begin{aligned} &P(E_1)=\frac{5}{9} \qquad ; \qquad P(E_2)=\frac{4}{9}\\ \end{aligned}

Total balls in 2nd bag = 10

\begin{aligned} &P\left (\frac{A}{E_1} \right )=\frac{6}{10} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{7}{10} \end{aligned}

Using total probabilty theorem

\begin{aligned} &\text { required probability }=P(A)=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )\\ &=\frac{5}{9}\times \frac{6}{10}+\frac{4}{9}\times \frac{7}{10}\\ &=\frac{1}{3}+\frac{14}{45}\\ &=\frac{15+14}{45}=\frac{29}{45} \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

Latest Question