What is the probability that 3 marbles drawn from 12 marbles are of different colours(with replacement) ?

July 28th, 2009 | by admin |

A box contains 3 white, 4 red and 5 black marbles. 3 marbles are drawn at random. Suppose the 3 marbles are drawn one at a time with replacement.
(a) What is the probability that they are of different colours ?
(b) What is the probability that they are of the same colour ?

Different Colors
P(White ∩ Red ∩ Black)
= [C(3,1)/C(12,1 x C(4,1)/C(12,1) x C(5,1)/C(12,1)] x P(3,3)
= [3/12 x 4/12 x 5/12] x 3!
= [3x4x5/12^3] x 3!

Same Color
P(All White U All Red U All Black)
= P(W1 ∩ W2 ∩ W3)
+ P(R1 ∩ R2 ∩ R3)
+ P(B1 ∩ B2 ∩ B3)
= [C(3,1)/C(12,1)]^3 + [C(4,1)/C(12,1)]^3 + [C(5,1)/C(12,1)]^3
= (3/12)^3 + (4/12)^3 + (5/12)^3
= 3^3 + 4^3 + 5^3/12^3

http://futureaccountant.com/probability/

  1. 3 Responses to “What is the probability that 3 marbles drawn from 12 marbles are of different colours(with replacement) ?”

  2. By Tatsuya on Jul 28, 2009 | Reply

    a) 35%
    b) 65%
    References :

  3. By krishbhavara on Jul 28, 2009 | Reply

    Different Colors
    P(White ∩ Red ∩ Black)
    = [C(3,1)/C(12,1 x C(4,1)/C(12,1) x C(5,1)/C(12,1)] x P(3,3)
    = [3/12 x 4/12 x 5/12] x 3!
    = [3x4x5/12^3] x 3!

    Same Color
    P(All White U All Red U All Black)
    = P(W1 ∩ W2 ∩ W3)
    + P(R1 ∩ R2 ∩ R3)
    + P(B1 ∩ B2 ∩ B3)
    = [C(3,1)/C(12,1)]^3 + [C(4,1)/C(12,1)]^3 + [C(5,1)/C(12,1)]^3
    = (3/12)^3 + (4/12)^3 + (5/12)^3
    = 3^3 + 4^3 + 5^3/12^3

    http://futureaccountant.com/probability/
    References :
    http://futureaccountant.com/probability/

  4. By Ahamanaku on Jul 28, 2009 | Reply

    a) 5/144
    b)
    References :

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