Login/Sign Up
Home » A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
CBSE | Class 12 | Maths | NCERT | Probability
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution:

Let \mathrm{E}_{1} be the event of bag selection. The event of choosing the bag, say bag II, and the event of drawing a red ball are represented by \mathrm{I}, \mathrm{E}_{2} and \mathrm{A}, respectively.

Then P\left(E_{1}\right)=P\left(E_{2}\right)=1 / 2

Also \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P} (drawing a red ball from bag I) = 4 / 8=1 / 2

And \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P} (drawing a red ball from bag II) = 2 / 8=1 / 4

Given that the ball in bag 1 is red, the probability of drawing it is P\left(E_{1} \mid A\right).

By using Bayes’ theorem, we have:

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}

=\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{4}}

=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}

=\frac{\frac{1}{4}}{\frac{3}{8}}=\frac{2}{3}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{2}{3}

i

More Material