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A bag contains contains 20 blue marbles, 20 green marbles, and 20 red marbles.

Two marbles are drawn one after the other. no replacement

Using selection (combinations), find the probability of:

(a) drawing two marbles of the same colour

(b) drawing a blue and a green marble

(c) drawing a blue or a green marble


Would the sample space be P(60,2) or C(60,2)

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  • $\begingroup$ You can use either, but best is to choose $C(60,2)$... $\endgroup$ – Nicolas FRANCOIS Dec 14 '16 at 1:46
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The question asks for combinations, but permutation seems more intuitive to me in this question, as the order of the marbles being drawn has to be taken into account.

I would use P(60,2) as the sample space. By doing so, I have to treat all 60 marbles as distinctive:
Blue: Marble 1, Marble 2, Marble 3..., Marble 20
Green: Marble 21, Marble 22, Marble 23..., Marble 40
Red: Marble 41 to Marble 60

(a) drawing two marbles of the same colour

$Probability = \frac{20P2+20P2+20P2}{60P2}$

(b) drawing a blue and a green marble

$Probability = \frac{(20P1*20P1)+(20P1*20P1)}{60P2}$

(c) drawing a blue or a green marble

$Probability = \frac{(20P1*40P1)+(20P1*40P1)}{60P2}$

Look forward to an answer that uses combinations instead.

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Maybe not exactly what is expected, but use for verification:

a) second is 19/59 chance of being same color

b) first has 40/60 chance of being successful, second has 20/59 chance of being the other required color. Multiply.

c) Anything except 2 red, so 1 - (20/60)*(19/59)

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