Question

Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is red is

• A. $\frac{19}{55}$
• B. $\frac{32}{55}$
• C. $\frac{41}{55}$
• D. $\frac{9}{55}$

Answer 1

Answer: (B)

$\frac{32}{55}$

There are four scenarios that can occur where final draw gives us red ball

Scenario 1

1st red drawn from urn a put into urn b

then black drawn from urn b put into urn a

then red drawn from urn a

Probability = $\frac{6}{10}\times \frac{6}{11}\times \frac{5}{10}=\frac{180}{1100}$

Scenario 2

1st black drawn from urn a put into urn b

then black drawn from urn b put into urn a

then red drawn from urn a

Probability = $\frac{4}{10}\times \frac{7}{11}\times \frac{6}{10}=\frac{168}{1100}$

Scenario 3

1st red drawn from urn a put into urn b

then red drawn from urn b put into urn a

then red drawn from urn a

Probability = $\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}=\frac{180}{1100}$

Scenario 4

1st black drawn from urn a put into urn b

then red drawn from urn b put into urn a

then red drawn from urn a

Probability = $\frac{4}{10}\times \frac{4}{11}\times \frac{7}{10}=\frac{112}{1100}$

Total probability$=\frac{640}{1100}=\frac{32}{55}$