A) \[\frac{8}{17}\]
B) \[\frac{40}{153}\]
C) \[\frac{5}{9}\]
D) \[\frac{4}{9}\]
Correct Answer: D
Solution :
Let a white ball be transferred from the first bag to the second. The Probability of selecting a white ball from the first bag \[=\frac{5}{9}.\] Now the second bag has 8 white and 9 black. The probability of selecting white ball from the second bag\[=\frac{8}{17}\]. Hence required probability \[=\frac{5}{9}\times \frac{8}{17}=\frac{40}{153}\] If a black ball be transferred from the first bag to the second, then the probability \[=\frac{4}{9}\times \frac{7}{17}=\frac{28}{153}\] Therefore required probability \[=\frac{40}{153}+\frac{28}{153}=\frac{4}{9}.\]
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