EAMCET Engineering 2010 Solved Paper

Case Ist Let a white ball be transfered from the first bag to the second bag.The probability of selecting white ball from the first bag is $P_1=\frac{3}{3+5}=\frac{3}{8}$Now, the second bag has 7 white and 8 black balls. The probability of selecting a white ball from the second bag is $P_2=\frac{7}{7+8}=\frac{7}{15}$The probability that both these events take place simultaneously$=P_1 \times P_2=\frac{3}{8} \times \frac{7}{15}=\frac{7}{40}$Case IInd Let a black ball be transfered from the first bag to the second bag.Its probability is $P_3=\frac{5}{3+5}=\frac{5}{8}$Now, the second bag contains 6 white and 9 black balls. The probability of drawing a white ball from the second bag is $P_4=\frac{6}{6+9}=\frac{6}{15}$$\therefore$ The probability of both these events taking place simultaneously$=P_3 \times P_4=\frac{5}{8} \times \frac{6}{15}=\frac{1}{4}$$\therefore$ The required probability $=P_1 P_2+P_3 P_4$$=\frac{7}{40}+\frac{1}{4}$$=\frac{7+10}{40}=\frac{17}{40}$