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Question

# An urn contains marbles of four colours : red, white, blue and green. When four marbles are drawn without replacement, the following events are equally likely : (1) the selection of four red marbles; (2) the selection of one white and three red marbles; (3) the selection of one white, one blue and two red marbles; (4) the selection of one marble of each colour. The smallest total number of marbles satisfying the given condition is

A
19
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B
21
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C
46
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D
69
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Solution

## The correct option is B 21Let number of red, white, blue, green balls be R,W,B,G respectively and R+W+B+G=n. Given that four marbles are drawn without replacement, and events are equally likely RC4nC4=WC1RC3nC4=WC1BC1RC2nC4=WC1BC1RC1GC1nC4 ⇒R(R−1)(R−2)(R−3)24=W.R.(R−1)(R−2)6=W.B.R(R−1)2=W.B.R.G From above expression we get, R=4W+3=2+3B=1+2G Now L.C.M of (4,3,2)=12 For Rmin=11⇒Wmin=2⇒Bmin=3⇒Gmin=5 ∴(R+W+B+G)min=21

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