Get Answers to all your Questions

header-bg qa

There are two bags, I and II. Bag I contains 3 red and 5 black balls and Bag II contains 4 red and 3 black balls. One ball is transferred randomly from Bag I to Bag II and then a ball is drawn randomly from Bag II. If the ball so drawn is found to be black in colour, then find the probability that the transferred ball is also black.

 

 

Answers (1)

\\ \mathrm{E}_{1}: \text {Ball transfered from bag I is black } \\ \mathrm{E}_{2}:\text {Ball transfered from bag I is red } \\ \mathrm{A}:\text { Ball drawn from bag II is black } \\

\\\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{5}{8} \\\\ \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{3}{8} \\ \\ \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)=\frac{4}{8}=\frac{1}{2} \\\\ \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)=\frac{3}{8}

\\ P\left(\frac{E_{1}}{A}\right)=\frac{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\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{\frac{5}{8} \cdot \frac{1}{2}}{\frac{5}{8} \cdot \frac{1}{2}+\frac{3}{8} \cdot \frac{3}{8}}\\\\=\frac{20}{29}

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads