Question

Bag $A$ contains $6$ red and $4$ black balls and bag $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from bag $B$ and placed in bag $A$. If one ball is now drawn from bag $A$. The probability that it is found to be red is

• A. $\frac{32}{55}$
• B. $\frac{33}{55}$
• C. $\frac{32}{63}$
• D. $\frac{25}{66}$

Answer 1

Answer: (A)

$\frac{32}{55}$

Bag A$\to 6$ Red$+4$ black

Bag B$\to 4$ Redd$+6$ black

$P_{AR}=\frac{6}{10};P_{AB}=\frac{4}{10};P_{BR}=\frac{4}{10};P_{BB}=\frac{6}{10}$

Let ball drawn from bag B is red

Now, $P_{A{R}^{\prime }}=\frac{7}{11}$ & $P_{A{B}^{\prime }}=\frac{4}{11}$

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

If ball drawn from bag B is black

$P_{A{R}^{\prime }}=\frac{6}{11};P_{A{B}^{\prime }}=\frac{5}{11}$

$\therefore$Probability$=\frac{6}{10}\times \frac{6}{11}$

Total probability$=\frac{28+36}{110}=\frac{64}{110}=\frac{32}{55}$