Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

A bag contains 4 red balls and 4 black balls. Another bag contains 2 red and 6 black. One of the 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.

Can you answer this question?

1 Answer

0 votes
  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability P(E_i|A) for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Step 1:
Let $E_1$ be the event that the first bag is selected, and $E_2$ the event that the second bag is selected.
Since $E_1$ and $E_2$ are mutually exclusive and exhaustive,
$P(E_1) = P(E_2) = \large\frac{1}{2}$
Let A be the event of drawing a red ball.
Given that Bag1 has 4 red and 4 black balls, n(Bag1) = 8, and n(R1) = 4.
Similarly for Bag2, n(Bag2) = 8 and n(R2) = 2.
$\Rightarrow$ P (drawing a red from Bag 1) = $\large \frac{4}{8} = \frac{1}{2}$ =$ P (A/E_1)$
$\Rightarrow$ P (drawing a red from Bag 2) = $\large \frac{2}{8} = \frac{1}{4}$ = $P(A/E_2)$
Step 2:
To find the probability of that the ball is drawn from the first bag, let's use Bayes theorem:
\(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Therefore the probability that the ball is drawn from the first bag = $P(E_1/A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
$P(E_1)(P(A/E_1) = \large\frac{1}{2}$$\times$$ \large \frac{1}{2} = \frac{1}{4}$
$P(E_1)P(A/E_1) + P(E_2)+P(A/E_2) = \large\frac{1}{2}$$\times$$\large\frac{1}{2} + \frac{1}{2}$$\times \large\frac{1}{4}$
$\qquad\qquad\qquad\qquad\qquad\qquad\;\;= \large\frac{2+1}{8}$
$\qquad\qquad\qquad\qquad\qquad\qquad\;\;=\large \frac{3}{8}$
Therefore $P(E_1/A) = \large \frac{\Large \frac{1}{4}}{\Large \frac{3}{8}}$$ =\large \frac{2}{3}$
answered Oct 3, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App