Question

Box $A$ contains $3$ red and $2$ black balls. Box $B$ contains $2$ red and $3$ black balls. One ball is drawn at random from box$A$ and placed in box $B$. Then one ball is drawn at random from the box $B$ and placed in $A$. The probability that the composition of balls in the two boxes remains unaltered is

• A. $\frac{9}{30}$
• B. $\frac{4}{15}$
• C. $\frac{17}{30}$
• D. $\frac{16}{30}$

Answer 1

The correct option is C $\frac{17}{30}$

Case1:

Let it is drawn from A and then from B also red it drawn and A to B.

P(red drawn from A):$\frac{{}^3{C}_1}{{}^5{C}_1}=\frac{3}{5}$

P(red again drawn from B)=$\frac{{}^3{C}_1}{{}^6{C}_1}=\frac{3}{6}$

P(total in this )=$\frac{3}{5}\times \frac{3}{6}=\frac{9}{30}$

Case 2 :

Same as case one but black

P(black drawn A)=$\frac{{}^2{C}_1}{{}^5{C}_1}=\frac{2}{5}$

P(black drawn B)=$\frac{{}^4{C}_1}{{}^5{C}_1}=\frac{4}{5}$

Total P=$\frac{2}{5}\times \frac{4}{6}=\frac{8}{30}$

So total=$\frac{9}{30}\times \frac{8}{30}=\frac{17}{30}$