Porisma

Worked solution

Use order counts instead of expanding every draw path

A box contains 5 white balls,3 red balls and 2 green balls.

(i) Three balls are taken at random with replacement. Find the probability that they all have different colours.

(ii) Five balls are taken at random without replacement. Find the probability that (a) exactly 2 are white, (b) at least 2 are white, (c) exactly 2 are red, given that at least 2 are white.

[SP2006/II/6]

CJC · Lecture · Example 10

(i) Method 1: sum the six different-colour paths

Let Wi,Ri,GiW_i,R_i,G_i denote the colour on draw i. With replacement the probabilities stay 0.5,0.3,0.2 at every split. The complete tree has 27 colour paths. The favourable paths are WRG, WGR, RWG, RGW, GWR, GRW, with products 0.5×0.3×0.2,0.5×0.2×0.3,0.3×0.5×0.2,0.3×0.2×0.5,0.2×0.5×0.3,0.2×0.3×0.5. Each is 0.03, so their sum is 0.18. The fifth symbolic term in both copies prints G1∩W1∩R2G_1\cap W_1\cap R_2; the intended GWR path is G1∩W2∩R3G_1\cap W_2\cap R_3, consistent with its numerical product.

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(i) Method 2: count the colour orders

There are 3! orders of three distinct colours, so P(all different)=P(W1∩R2∩G3) 3!=0.5×0.3×0.2×6=0.18P(\text{all different})=P(W_1\cap R_2\cap G_3)\,3!=0.5\times0.3\times0.2\times6=0.18. Follow the Teacher guidance: the full tree is tedious and unnecessary for this calculation; use the order count once the one-path product is clear.

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(ii) Choose a tractable model without replacement

For five draws the source describes a 3^5=243-case full colour tree. This counts candidate colour strings, including impossible strings beyond the available colour supplies; it is not 243 equally likely feasible outcomes. The source therefore uses order counts or combinations instead. Probabilities must now update after each draw.

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(ii)(a) Method 1: count the orders of WWNNN

Write N for non-white. The specified order WWNNN has probability (5/10)(4/9)(5/8)(4/7)(3/6)(5/10)(4/9)(5/8)(4/7)(3/6). There are 5!/(2!3!)=105!/(2!3!)=10 positions for the two white and three non-white results. Multiplying gives P(exactly two white)=25/63P(\text{exactly two white})=25/63.

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(ii)(b) Method 1: subtract zero and one white

The no-white probability is (5/10)(4/9)(3/8)(2/7)(1/6)(5/10)(4/9)(3/8)(2/7)(1/6). One specified WNNNN order has probability (5/10)(5/9)(4/8)(3/7)(2/6)(5/10)(5/9)(4/8)(3/7)(2/6), with 5!/4!=55!/4!=5 orders. Subtract the first probability and five times the second from 1. Thus P(at least two white)=113/126P(\text{at least two white})=113/126. This implements the Teacher complement annotation.

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(ii)(c) Method 1: split the favourable intersection

Two red and at least two white among five balls means RRWWG or RRWWW. One RRWWG order has probability (3/10)(2/9)(5/8)(4/7)(2/6)(3/10)(2/9)(5/8)(4/7)(2/6) and 5!/(2!2!)=305!/(2!2!)=30 orders. One RRWWW order has probability (3/10)(2/9)(5/8)(4/7)(3/6)(3/10)(2/9)(5/8)(4/7)(3/6) and 5!/(2!3!)=105!/(2!3!)=10 orders. The divisors correct the two repeated red results and either two or three repeated white results.

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(ii)(c) Divide the intersection probability by the condition

Add 30 times the RRWWG probability and 10 times the RRWWW probability to obtain P(two red and at least two white)=5/14P(\text{two red and at least two white})=5/14. Divide by 113/126 from (b): (5/14)/(113/126)=45/113(5/14)/(113/126)=45/113.

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(ii) Method 2: label the ten physical balls

Use W1,W2,W3,W4,W5,R1,R2,R3,G1,G2 as physical identities, distinct from draw-index events. There are (105)=252\binom{10}{5}=252 equally likely unordered five-ball selections.

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(ii)(a) Method 2: select white and non-white balls

Choose 2 of the 5 white and 3 of the 5 non-white balls: (52)(53)=100\binom52\binom53=100. The probability is 100/252=25/63. Both selection factors and their roles are retained.

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(ii)(b) Method 2: remove the two excluded counts

Zero white contributes (55)=1\binom55=1 selection of all five non-white balls. Exactly one white contributes (51)(54)=25\binom51\binom54=25: choose 1 white from 5 and 4 non-white from 5. Therefore 1−1/252−25/252=226/252=113/1261-1/252-25/252=226/252=113/126, retaining the Teacher-completed counts.

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(ii)(c) Method 2: count both intersection cases

RRWWG contributes (32)(52)(21)=60\binom32\binom52\binom21=60: choose 2 red from 3,2 white from 5,1 green from 2. RRWWW contributes (32)(53)=30\binom32\binom53=30: choose 2 red from 3 and 3 white from 5. The conditional probability is [(60+30)/252]/(113/126)=45/113[(60+30)/252]/(113/126)=45/113. Equivalently 90 favourable selections lie among 226 eligible selections.

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Compare your sketch

Three colour draws with replacement

Expand a first-draw branch to read its nine complete paths. All 27 ordered outcomes are retained.

Start · probability 1

Open a branch to follow its outcomes. Outcome percentages are for complete paths. Open an outcome to see its calculation.

  • W₁ · White0.5 · 9 outcomes

    W₁ · White

    • W₂ · White0.5 · 3 outcomes

      W₁ · White → W₂ · White

      • W₃ · WhitePath 12.5%
        W₁ · White → W₂ · White → W₃ · White
        Branch weights
        0.5 × 0.5 × 0.5
        Complete path probability
        1/8= 0.125 · 12.5%
      • R₃ · RedPath 7.5%
        W₁ · White → W₂ · White → R₃ · Red
        Branch weights
        0.5 × 0.5 × 0.3
        Complete path probability
        3/40= 0.075 · 7.5%
      • G₃ · GreenPath 5%
        W₁ · White → W₂ · White → G₃ · Green
        Branch weights
        0.5 × 0.5 × 0.2
        Complete path probability
        1/20= 0.05 · 5%
    • R₂ · Red0.3 · 3 outcomes

      W₁ · White → R₂ · Red

      • W₃ · WhitePath 7.5%
        W₁ · White → R₂ · Red → W₃ · White
        Branch weights
        0.5 × 0.3 × 0.5
        Complete path probability
        3/40= 0.075 · 7.5%
      • R₃ · RedPath 4.5%
        W₁ · White → R₂ · Red → R₃ · Red
        Branch weights
        0.5 × 0.3 × 0.3
        Complete path probability
        9/200= 0.045 · 4.5%
      • G₃ · GreenPath 3% · All different
        W₁ · White → R₂ · Red → G₃ · Green
        Branch weights
        0.5 × 0.3 × 0.2
        Complete path probability
        3/100= 0.03 · 3%
        All different
    • G₂ · Green0.2 · 3 outcomes

      W₁ · White → G₂ · Green

      • W₃ · WhitePath 5%
        W₁ · White → G₂ · Green → W₃ · White
        Branch weights
        0.5 × 0.2 × 0.5
        Complete path probability
        1/20= 0.05 · 5%
      • R₃ · RedPath 3% · All different
        W₁ · White → G₂ · Green → R₃ · Red
        Branch weights
        0.5 × 0.2 × 0.3
        Complete path probability
        3/100= 0.03 · 3%
        All different
      • G₃ · GreenPath 2%
        W₁ · White → G₂ · Green → G₃ · Green
        Branch weights
        0.5 × 0.2 × 0.2
        Complete path probability
        1/50= 0.02 · 2%
  • R₁ · Red0.3 · 9 outcomes

    R₁ · Red

    • W₂ · White0.5 · 3 outcomes

      R₁ · Red → W₂ · White

      • W₃ · WhitePath 7.5%
        R₁ · Red → W₂ · White → W₃ · White
        Branch weights
        0.3 × 0.5 × 0.5
        Complete path probability
        3/40= 0.075 · 7.5%
      • R₃ · RedPath 4.5%
        R₁ · Red → W₂ · White → R₃ · Red
        Branch weights
        0.3 × 0.5 × 0.3
        Complete path probability
        9/200= 0.045 · 4.5%
      • G₃ · GreenPath 3% · All different
        R₁ · Red → W₂ · White → G₃ · Green
        Branch weights
        0.3 × 0.5 × 0.2
        Complete path probability
        3/100= 0.03 · 3%
        All different
    • R₂ · Red0.3 · 3 outcomes

      R₁ · Red → R₂ · Red

      • W₃ · WhitePath 4.5%
        R₁ · Red → R₂ · Red → W₃ · White
        Branch weights
        0.3 × 0.3 × 0.5
        Complete path probability
        9/200= 0.045 · 4.5%
      • R₃ · RedPath 2.7%
        R₁ · Red → R₂ · Red → R₃ · Red
        Branch weights
        0.3 × 0.3 × 0.3
        Complete path probability
        27/1000= 0.027 · 2.7%
      • G₃ · GreenPath 1.8%
        R₁ · Red → R₂ · Red → G₃ · Green
        Branch weights
        0.3 × 0.3 × 0.2
        Complete path probability
        9/500= 0.018 · 1.8%
    • G₂ · Green0.2 · 3 outcomes

      R₁ · Red → G₂ · Green

      • W₃ · WhitePath 3% · All different
        R₁ · Red → G₂ · Green → W₃ · White
        Branch weights
        0.3 × 0.2 × 0.5
        Complete path probability
        3/100= 0.03 · 3%
        All different
      • R₃ · RedPath 1.8%
        R₁ · Red → G₂ · Green → R₃ · Red
        Branch weights
        0.3 × 0.2 × 0.3
        Complete path probability
        9/500= 0.018 · 1.8%
      • G₃ · GreenPath 1.2%
        R₁ · Red → G₂ · Green → G₃ · Green
        Branch weights
        0.3 × 0.2 × 0.2
        Complete path probability
        3/250= 0.012 · 1.2%
  • G₁ · Green0.2 · 9 outcomes

    G₁ · Green

    • W₂ · White0.5 · 3 outcomes

      G₁ · Green → W₂ · White

      • W₃ · WhitePath 5%
        G₁ · Green → W₂ · White → W₃ · White
        Branch weights
        0.2 × 0.5 × 0.5
        Complete path probability
        1/20= 0.05 · 5%
      • R₃ · RedPath 3% · All different
        G₁ · Green → W₂ · White → R₃ · Red
        Branch weights
        0.2 × 0.5 × 0.3
        Complete path probability
        3/100= 0.03 · 3%
        All different
      • G₃ · GreenPath 2%
        G₁ · Green → W₂ · White → G₃ · Green
        Branch weights
        0.2 × 0.5 × 0.2
        Complete path probability
        1/50= 0.02 · 2%
    • R₂ · Red0.3 · 3 outcomes

      G₁ · Green → R₂ · Red

      • W₃ · WhitePath 3% · All different
        G₁ · Green → R₂ · Red → W₃ · White
        Branch weights
        0.2 × 0.3 × 0.5
        Complete path probability
        3/100= 0.03 · 3%
        All different
      • R₃ · RedPath 1.8%
        G₁ · Green → R₂ · Red → R₃ · Red
        Branch weights
        0.2 × 0.3 × 0.3
        Complete path probability
        9/500= 0.018 · 1.8%
      • G₃ · GreenPath 1.2%
        G₁ · Green → R₂ · Red → G₃ · Green
        Branch weights
        0.2 × 0.3 × 0.2
        Complete path probability
        3/250= 0.012 · 1.2%
    • G₂ · Green0.2 · 3 outcomes

      G₁ · Green → G₂ · Green

      • W₃ · WhitePath 2%
        G₁ · Green → G₂ · Green → W₃ · White
        Branch weights
        0.2 × 0.2 × 0.5
        Complete path probability
        1/50= 0.02 · 2%
      • R₃ · RedPath 1.2%
        G₁ · Green → G₂ · Green → R₃ · Red
        Branch weights
        0.2 × 0.2 × 0.3
        Complete path probability
        3/250= 0.012 · 1.2%
      • G₃ · GreenPath 0.8%
        G₁ · Green → G₂ · Green → G₃ · Green
        Branch weights
        0.2 × 0.2 × 0.2
        Complete path probability
        1/125= 0.008 · 0.8%
Compare outcome probabilities
Multiply along each complete path. All outcomes together have probability 1.
OutcomeProbability
W₁ · White → W₂ · White → W₃ · White1/8= 0.125 · 12.5%
W₁ · White → W₂ · White → R₃ · Red3/40= 0.075 · 7.5%
W₁ · White → W₂ · White → G₃ · Green1/20= 0.05 · 5%
W₁ · White → R₂ · Red → W₃ · White3/40= 0.075 · 7.5%
W₁ · White → R₂ · Red → R₃ · Red9/200= 0.045 · 4.5%
W₁ · White → R₂ · Red → G₃ · Green3/100= 0.03 · 3%
W₁ · White → G₂ · Green → W₃ · White1/20= 0.05 · 5%
W₁ · White → G₂ · Green → R₃ · Red3/100= 0.03 · 3%
W₁ · White → G₂ · Green → G₃ · Green1/50= 0.02 · 2%
R₁ · Red → W₂ · White → W₃ · White3/40= 0.075 · 7.5%
R₁ · Red → W₂ · White → R₃ · Red9/200= 0.045 · 4.5%
R₁ · Red → W₂ · White → G₃ · Green3/100= 0.03 · 3%
R₁ · Red → R₂ · Red → W₃ · White9/200= 0.045 · 4.5%
R₁ · Red → R₂ · Red → R₃ · Red27/1000= 0.027 · 2.7%
R₁ · Red → R₂ · Red → G₃ · Green9/500= 0.018 · 1.8%
R₁ · Red → G₂ · Green → W₃ · White3/100= 0.03 · 3%
R₁ · Red → G₂ · Green → R₃ · Red9/500= 0.018 · 1.8%
R₁ · Red → G₂ · Green → G₃ · Green3/250= 0.012 · 1.2%
G₁ · Green → W₂ · White → W₃ · White1/20= 0.05 · 5%
G₁ · Green → W₂ · White → R₃ · Red3/100= 0.03 · 3%
G₁ · Green → W₂ · White → G₃ · Green1/50= 0.02 · 2%
G₁ · Green → R₂ · Red → W₃ · White3/100= 0.03 · 3%
G₁ · Green → R₂ · Red → R₃ · Red9/500= 0.018 · 1.8%
G₁ · Green → R₂ · Red → G₃ · Green3/250= 0.012 · 1.2%
G₁ · Green → G₂ · Green → W₃ · White1/50= 0.02 · 2%
G₁ · Green → G₂ · Green → R₃ · Red3/250= 0.012 · 1.2%
G₁ · Green → G₂ · Green → G₃ · Green1/125= 0.008 · 0.8%
All 27 outcomes1= 1 · 100%
Tree assumptions and source notes

W: white, R: red, G: green. Every split uses 0.5, 0.3 and 0.2 because each ball is replaced.

Draw indices 1, 2 and 3 identify events, not individual balls.

The six all-different paths are WRG, WGR, RWG, RGW, GWR and GRW. They are labeled All different in the tree; each has probability 0.03.

This diagram is for the with-replacement experiment in part (i), not the five-ball selection without replacement in part (ii).

Open original probability tree (new tab)
All 27 ordered colour paths are retained. Each of three stages branches to white W with probability 0.5, red R with probability 0.3, or green G with probability 0.2, independent of preceding colours because draws are with replacement. Nodes have draw indices 1, 2 and 3. The six all-different paths WRG, WGR, RWG, RGW, GWR and GRW are marked by orange final branches; path labels are supplementary. Every first, second and third-stage conditional branch probability is shown.

Common mistakes

Part (i) uses replacement; part (ii) does not. Keep order-count factors with sequential probabilities and combination counts with the common equally likely selection space. The original symbolic-index slip is qualified rather than propagated.

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