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WAEC 2011 · Paper 2 · Q15

A bag contains 4 red, 6 blue and 8 green identical marbles.

  1. (a)

    If three marbles are drawn at random, without replacement, calculate the probability that: (i) all will be green; (ii) all will have the same colour.

    Separate values with commas, e.g. 3, −2

  2. (b)

    If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.

Worked solution (try it first)
  1. There are 18 marbles: 4 red, 6 blue and 8 green.

(a)(i)

  1. Without replacement: 818×717×616=3364896\dfrac{8}{18} \times \dfrac{7}{17} \times \dfrac{6}{16} = \dfrac{336}{4896}
    =7102= \dfrac{7}{102}.

(ii)

  1. All red: 4×3×24896=244896\dfrac{4 \times 3 \times 2}{4896} = \dfrac{24}{4896}.
  2. All blue: 6×5×44896=1204896\dfrac{6 \times 5 \times 4}{4896} = \dfrac{120}{4896}.
  3. Add the three: 24+120+3364896=4804896\dfrac{24 + 120 + 336}{4896} = \dfrac{480}{4896}
    =551= \dfrac{5}{51}
    ≈0.098\approx 0.098.

(b)

  1. With replacement: (418)3+(618)3+(818)3=64+216+5125832\left(\dfrac{4}{18}\right)^3 + \left(\dfrac{6}{18}\right)^3 + \left(\dfrac{8}{18}\right)^3 = \dfrac{64 + 216 + 512}{5832}
    =1181= \dfrac{11}{81}
    ≈0.136\approx 0.136.

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