Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • There are two boxes I and II. Box I contains 3 red and 6 black balls. Box II contains 5 red and 'n' black balls. One of the two boxes,box I and box II is selscted at random and a ball is drawn at random. The ball drawn is found to be red

There are two boxes I and II. Box I contains 3 red and 6 black balls. Box II contains 5 red
and 'n' black balls. One of the two boxes,box I and box II is selscted at random and a ball is drawn at random. The ball drawn is found to be red. If The probability that this red ball comes out from box II is \frac{3}{5}, find the value of 'n'.

 

 

 

 
 
 
 
 

Answers (1)

E1 =  selecting box I
E2 = selecting box II
A = geting a red ball from selescted box
p\left ( E_{1} \right )= \frac{1}{2} , \: \: p\left ( E_{1} \right )= \frac{1}{2}
p\left ( \frac{A}{E_{1}} \right )= \frac{3}{9}= \frac{1}{3}
p\left ( \frac{A}{E_{2}} \right )= \frac{p\left ( E_{2} \right )\cdot p\left ( \frac{A}{E_{2}} \right )}{p\left ( E_{1} \right )\cdot p\left ( \frac{A}{E_{1}} \right )+p\left ( E_{2} \right )\cdot p\left ( \frac{A}{E_{2}} \right )}
\frac{3}{5}= \frac{\frac{1}{2}\times \frac{5}{n+5}}{\frac{1}{2}\times \frac{1}{3}+\frac{1}{2}\times \frac{5}{n+5}}
\frac{3}{5}= \frac{15}{n+20}
\left ( n+20 \right )3= 75
3n= 15
n= 3

Posted by

Ravindra Pindel

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 10th, 12th Results 2025 Live: How to check CBSE result on Digilocker? Date, time updates
anam-khan

Board Exam Result 2025 Live: UK Board 10th, 12th results tomorrow; AP SSC, TS Inter, UP, CBSE updates

Latest Question