A box contains 9 whites, 7 red and 4 blue balls. Three balls are picked at random, one after the other without replacement. a) find the probability that: i. they are all of the same color ii. there is one of each colour iii. two of them are red b) if it is known that the first one picked is blue, find the probability that the rest are white.


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Question Number 56678 by pieroo last updated on 21/Mar/19

$A box contains 9 whites, 7 red and 4 blue$ $balls . Three balls are picked at random,$ $one after the other without replacement .$ $a) find the probability that :$ $i . they are all of the same color$ $ii . there is one of each colour$ $iii . two of them are red$ $b) if it is known that the first one picked$ $is blue, find the probability that the rest$ $are white .$
	Answered by MJS last updated on 21/Mar/19

	$a . i .$ $p (www) = \frac{9 \times 8 \times 7}{20 \times 19 \times 18} = \frac{7}{95}$ $p (rrr) = \frac{7 \times 6 \times 5}{20 \times 19 \times 18} = \frac{7}{228}$ $p (bbb) = \frac{4 \times 3 \times 2}{20 \times 19 \times 18} = \frac{1}{285}$ $\frac{7}{95} + \frac{7}{228} + \frac{1}{285} = \frac{41}{380} \approx 10 . 79 %$ $a . ii .$ $p (wrb) = p (wbr) = p (rwb) = . . . 6 equal possibilities$ $6 \times \frac{9 \times 7 \times 4}{20 \times 19 \times 18} = \frac{21}{95} \approx 22 . 11 %$ $a . iii .$ $p (rrx) = p (rxr) = p (xrr) 3 equal possibilities$ $3 \times \frac{7 \times 6 \times 13}{20 \times 19 \times 18} = \frac{91}{380} \approx 23 . 95 %$ $b .$ $p (ww) = \frac{9 \times 8}{19 \times 18} = \frac{4}{19} \approx 21 . 05 %$
	Commented by pieroo last updated on 21/Mar/19

	$Thank you very very much, Sir .$
	Commented by malwaan last updated on 22/Mar/19

	$a . ii . 6 = 3!$
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