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'.

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

Related Chapters

Preparation Products

Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-