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Question

A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.

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Solution

Consider the given events.
A = A white ball in the first draw
B = A black ball in the second draw
C = A red ball in the third draw

Now, PA=416=14PB/A=715PC/AB=514 Required probability = PABC=PA×PB/A×PC/AB =14×715×514 =124

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