A) \[\frac{7}{16}\]
B) \[\frac{9}{32}\]
C) \[\frac{7}{8}\]
D) \[\frac{9}{16}\]
Correct Answer: A
Solution :
For the bag A we can see that there are 2 white, 3 red and 2 black balls. Similarly, from bag B we have 4 white, 2 red and 3 black balls. Probability of choosing a white and then a red ball from bag B is given by\[=\frac{^{4}{{C}_{1}}{{\times }^{2}}{{C}_{1}}}{^{9}{{C}_{2}}}\] Probability of choosing a white ball then a red ball from bag A is given by\[=\frac{^{2}{{C}_{1}}{{\times }^{3}}{{C}_{1}}}{^{7}{{C}_{2}}}\] So, the probability of getting a white ball and then a red ball from bag B is given by \[\frac{\frac{^{4}{{C}_{1}}{{\times }^{2}}{{C}_{1}}}{^{9}{{C}_{2}}}}{\frac{^{4}{{C}_{1}}{{\times }^{2}}{{C}_{1}}}{^{9}{{C}_{2}}}+\frac{^{2}{{C}_{1}}{{\times }^{3}}{{C}_{1}}}{^{7}{{C}_{2}}}}=\frac{\frac{2}{9}}{\frac{2}{7}+\frac{2}{9}}=\frac{2\times 7}{18+14}=\frac{7}{16}\]
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