Question

Urn A contains six red and four black balls and urn B has four red and six black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is transferred at random from urn B to urn A. If one ball is now drawn at random from urn A, find the probability that is red.

• A. $\frac{32}{65}$
• B. $\frac{32}{55}$
• C. $\frac{23}{55}$
• D. $\frac{56}{65}$

Answer 1

The correct option is B $\frac{32}{55}$

Let $E_{rr}$denote that a red colour ball is transferred from urn A to urn B then a red colour ball is transferred from urn B to urn A
$E_{rb}$ denote that a red colour ball is transferred from urn A to urn B then a black colour ball is transferred from urn B to urn A
$E_{br}$ denote that a black colour ball is transferred from urn A to urn B then a red colour ball is transferred from urn B to urn A.
$E_{bb}$ denote that a black colour ball is transferred from urn A to urn B then a black colour ball is transferred from urn B to urn A. Then
$P\left(E_{rr}\right)=\left(\frac{6}{10}\right)\frac{5}{11}=\frac{3}{11},$
$P\left(E_{rb}\right)=\left(\frac{6}{10}\right)\frac{6}{11}=\frac{18}{55},$
$P\left(E_{br}\right)=\left(\frac{4}{10}\right)\frac{4}{11}=\frac{8}{5},$
$P\left(E_{bb}\right)=\left(\frac{4}{10}\right)\left(\frac{7}{11}\right)=\frac{14}{55}$
Let A be the event of drawing a red colour ball after these transfers. Then
$P\left(\frac{A}{E_{rr}}\right)=\frac{6}{10},P\left(\frac{A}{E_{rb}}\right)=\frac{5}{10}P\left(\frac{A}{E_{br}}\right)=\frac{7}{10},P\left(\frac{A}{E_{bb}}\right)=\frac{6}{10}$
Therefore, the requird probability is
$P\left(A\right)=P\left(E_{rr}\right)P\left(\frac{A}{E_{rr}}\right)+P\left(E_{rb}\right)P\left(\frac{A}{E_{rb}}\right)$
$+P\left(E_{br}\right)P\left(\frac{A}{E_{br}}\right)+P\left(E_{b}b\right)P\left(\frac{A}{E_{bb}}\right)=\left(\frac{3}{11}\right)\left(\frac{6}{10}\right)+\left(\frac{5}{10}\right)\frac{18}{55}+\left(\frac{8}{55}\right)\left(\frac{7}{10}\right)+\left(\frac{14}{55}\right)\left(\frac{6}{10}\right)=\frac{90+90+56+84}{550}=\frac{32}{55}$