Suppose that a bag A contains n red and 2 black balls and another bag B contains 2 red and n black balls. One of the two bags is selected at random and two balls are drawn from it at a time. When it is known that the two balls drawn are red, if the probability that those two balls drawn are from bag A is 6/7, then n =

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

	A	B
Red	n	2
Black	2	n

$E_{1} =$ Bag A Selected $P (E_{1}) = \frac{1}{2}$

$E_{2} =$ Bag A Selected $P (E_{2}) = \frac{1}{2}$

A=two balls are drawn it is red

$\begin{array}{rcl} P (\frac{A}{E_{1}}) & = & \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}} \end{array}$

$\begin{array}{rcl} P (\frac{A}{E_{2}}) & = & \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}} \end{array}$

$\begin{array}{rcl} P (\frac{E_{1}}{A}) & = & \frac{\frac{1}{2} \times \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}}}{\frac{1}{2} \times \frac{{}^{n}{C_{2}}}{{}^{(n + 2)}{C_{2}}} + \frac{1}{2} \frac{{}^{2}{C_{2}}}{{}^{(n + 2)}{C_{2}}}} = \frac{6}{7} \end{array}$

$\begin{array}{rcl} \frac{{}^{n}{C_{2}}}{{}^{n}{C_{2}} + 1} & = & \frac{6}{7} \end{array}$

$\begin{array}{rcl} \Rightarrow & {}^{n}{C_{2}} = 6 \Rightarrow n = \end{array} 4$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET