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A bag contains 6 red and 9 blue balls. Two successive drawing of four balls are made such that the balls are not replaced before the second draw.Find the probability that the first draw gives 4 red balls and the second draw gives 4 blue balls.

(a) \(\frac{3}{715}\) 

(b) \(\frac{7}{715}\) 

(c) \(\frac{15}{233}\) 

(d) None of these

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Answer: (a) \(\frac{3}{715}\)

Let A: Event of drawing 4 red balls in the first draw and 

B: Event of drawing 4 blue balls in the second draw without replacement of the balls drawn in A.

Then, P(A) = \(\frac{n(A)}{n(S)}\)

=\(\frac{No.\, of\, ways\, of\, drawing\, 4\, red\, balls\, out\, of\, 6 \,red\, balls}{No.\, of \,ways\, of\, drawing\, 4\, red\, balls\, out\, of\, 6\, red\, balls}\)

\(P\big(\frac{B}{A}\big)\) = \(\frac{n(B)}{n(S)}\)

\(\frac{No.\, of\, ways\, of\, drawing\, 4\, blue\, balls\, out\, of\, 9\, blue\, balls}{No.\, of\, ways\, of\, drawing\, 4\, balls\, out\, of\, 11\, balls\, remaining\, after\, the\, 1st\, draw}\)

Note: This second event is denoted by B/A as it depends on condition A.

∴  Required probability = P(A) × P(B/A)

\(\frac{1}{91}\times \frac{21}{55} = \frac{3}{715}\)

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