If from each of the three boxes containing 3 blue and 1 red balls, 2 b

If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, 1 blue and 3 red balls, one ball is drawn at random, then the probability that 2 blue and 1 red ball will be drawn is :(a) \(\frac{13}{32}\)(b) \(\frac{27}{32}\)(c) \(\frac{19}{32}\)(d) None of these

Answer»

(a) \(\frac{13}{32}\)

The three boxes B₁, B₂ and B₂ contain the different coloured balls as follows :

	Blue	Red
Box 1	3	1
Box 2	2	2
Box 3	1	3

There can be three mutually exclusive cases of drawing 2 blue balls and 1 red balls in the ways as given :

	Box 1	Box 2	Box 3
Case I	1 Blue	1 Blue	1 Red
Case II	1 Blue	1 Red	1 Blue
Case III	1 Red	1 Blue	1 Blue

∴ P(Drawing 2 blue and 1 red ball)

= P(B₁ (Blue), B₂ (Blue), B₃ (Red)) + P(B₁ (Blue), B₂ (Red), B₃ (Blue)) + P(B₁ (Red), B₂ (Blue), B₃ (Blue))

= \(\frac{3}{4}\) x \(\frac{2}{4}\) x \(\frac{3}{4}\) + \(\frac{3}{4}\) x \(\frac{2}{4}\) x \(\frac{1}{4}\) + \(\frac{1}{4}\) x \(\frac{2}{4}\) x \(\frac{1}{4}\)

= \(\frac{18}{64}\) + \(\frac{6}{64}\) + \(\frac{2}{64}\) = \(\frac{26}{64}\) = \(\frac{13}{32}\).

If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, 1 blue and 3 red balls, one ball is drawn at random, then the probability that 2 blue and 1 red ball will be drawn is :(a) \(\frac{13}{32}\)(b) \(\frac{27}{32}\)(c) \(\frac{19}{32}\)(d) None of these

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