Question

Urn 1 contains two white balls and one black ball, while urn 2 contains one whiteball and ve black balls. One ball is drawn at random from urn 1 and placed in urn2. A ball is then drawn from urn 2. It happens to be white. What is the probabilitythat the transferred ball was white?

Answer 1

Answer:

4/5 is the probability of white ball from urn 2

Step-by-step explanation:

Correct statement of the question is ;

Urn 1 contains two white balls and one black ball, while urn 2 contains one white bait and five black balls. One ball is drawn at random from urn 1 and placed in urn 2. A ball is then drawn from urn 2. It happens to be white. What is the probability that the transferred ball was white?

We solve it ;

Probability of taking out balls from urn 1

a) for white ball

$P(w_{1} )=\frac{2}{3}$

 

b) for balck ball

$P(b_{1} )= \frac{1}{3}$

If one ball is taken from urn 1 and put in urn 2 (which readily contains one white and five black balls)

Probability of taking out balls from urn 2

a) for white ball

$P(w_{2} )=P(\frac{w_{2} }{w_{1} } ).P(w_{1} )+P(\frac{w_{2} }{b_{1} } ).P(b_{1} )$

$=\frac{2}{7} . \frac{2}{3} + \frac{1}{7}. \frac{1}{3} =\frac{5}{21}$

Then the probability that white ball was transferred from urn 1 to urn 2 is;

$P(\frac{white ball transfer }{w_{2} } )= \frac{P(\frac{w_{2} }{w_{1} } ).P(w_{2} )}{P(w_{2}) }$

$=\frac{ \frac{2}{7}.\frac{2}{3} }{\frac{5}{21} }$

$=\frac{4}{5}$

So, the probability that white ball is drawn from urn 2 is 4/5