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Worked solution

Retain the exceptional all-blue score in the conditional game

A box contains 2 red balls,3 green balls and x blue balls, with integer x≥5x\ge5. A contestant chooses 5 balls without replacement. S is the number of green balls plus three times the number of red balls. Blue balls score nothing unless all 5 balls are blue, in which case S=25.

(i) Show that P(S=6)=20x(x2−3x+14)(x+5)(x+4)(x+3)(x+2)(x+1)P(S=6)=\dfrac{20x(x^2-3x+14)}{(x+5)(x+4)(x+3)(x+2)(x+1)}. marks

(ii) Given P(S=6)=5/63P(S=6)=5/63, calculate x. marks

(iii) Complete the distribution table. marks

sGiven P(S=s)P(S=s)
1to find
25/425/42
35/635/63
4to find
55/215/21
65/635/63
7to find
85/845/84
91/2521/252
25to find

(iv) Evaluate E(S)E(S) and find P(S>E(S))P(S>E(S)). marks

(v) Find the probability that no green balls are drawn, given S>E(S)S>E(S). marks

[2017/CJC/II/10]

CJC · Review problems · Question 9

Answer

(i)formula shown; (ii)x=5; (iii)table above; (iv)E(S)=1159/252≈4.60E(S)=1159/252\approx4.60, upper-tail probability 127/252≈0.504127/252\approx0.504; (v)11/12711/127.

(i) Retain both score-six compositions and ordering factors

Write R,G,B for ball colours and D=(x+5)(x+4)(x+3)(x+2)(x+1)D=(x+5)(x+4)(x+3)(x+2)(x+1). Score 6 arises from RRBBB or RGGGB.

P(S=6)=2x+51x+4xx+3x−1x+2x−2x+15!3!2!+2x+53x+42x+31x+2xx+15!3!.\begin{aligned}&P(S=6)\\ & =\frac2{x+5}\frac1{x+4}\frac{x}{x+3}\frac{x-1}{x+2}\frac{x-2}{x+1}\frac{5!}{3!2!}\\ &{}+\frac2{x+5}\frac3{x+4}\frac2{x+3}\frac1{x+2}\frac{x}{x+1}\frac{5!}{3!}.\end{aligned}

This is 20x(x−1)(x−2)/D+240x/D=20x[(x−1)(x−2)+12]/D=20x(x2−3x+14)/D20x(x-1)(x-2)/D+240x/D=20x[(x-1)(x-2)+12]/D=20x(x^2-3x+14)/D.

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(ii) Retain the source equation and integer restriction

Equate 5/63=20x(x2−3x+14)/D5/63=20x(x^2-3x+14)/D. Thus 5(x+5)(x+4)(x+3)(x+2)(x+1)−1260x(x2−3x+14)=05(x+5)(x+4)(x+3)(x+2)(x+1)-1260x(x^2-3x+14)=0. Solving under the integer restriction x≥5x\ge5 gives x=5.

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(iii) Retain every missing-entry calculation

For x=5, there are 10 balls. The source sequential calculations include all distinct colour orderings:

P(S=1)=P(GBBBB)=(3/10)(5/9)(4/8)(3/7)(2/6)(5!/4!)=5/84P(S=1)=P(GBBBB)=(3/10)(5/9)(4/8)(3/7)(2/6)(5!/4!)=5/84. The factor counts arrangements of one G and four repeated Bs.

P(S=4)=P(RGBBB)=(2/10)(3/9)(5/8)(4/7)(3/6)(5!/3!)=5/21P(S=4)=P(RGBBB)=(2/10)(3/9)(5/8)(4/7)(3/6)(5!/3!)=5/21, arranging three repeated Bs.

P(S=7)=P(RRGBB)=(2/10)(1/9)(3/8)(5/7)(4/6)(5!/(2!2!))=5/42P(S=7)=P(RRGBB)=(2/10)(1/9)(3/8)(5/7)(4/6)(5!/(2!2!))=5/42, arranging two repeated Rs and two repeated Bs.

P(S=25)=P(BBBBB)=(5/10)(4/9)(3/8)(2/7)(1/6)=1/252P(S=25)=P(BBBBB)=(5/10)(4/9)(3/8)(2/7)(1/6)=1/252.

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(iii) Complete distribution

sP(S=s)P(S=s)
15/845/84
25/425/42
35/635/63
45/215/21
55/215/21
65/635/63
75/425/42
85/845/84
91/2521/252
251/2521/252
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(iv) Mean and upper-tail probability

Using the full table, E(S)=1159/252≈4.60E(S)=1159/252\approx4.60. Therefore S>E(S)S>E(S) selects scores 5,6,7,8,9,25, and its probability is 127/252≈0.504127/252\approx0.504. The source uses the rounded cutoff 4.60; the exact mean selects the same integer scores.

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(v) Retain both no-green upper-tail cases

No green balls with score above the mean means RRBBB or BBBBB. Thus P(no G∣S>E(S))=[P(RRBBB)+P(BBBBB)]/(127/252)=[20(5)(4)(3)/((10)(9)(8)(7)(6))+1/252]/(127/252)=[10/252+1/252]/(127/252)=11/127P(\text{no G}\mid S>E(S))=[P(RRBBB)+P(BBBBB)]/(127/252)=[20(5)(4)(3)/((10)(9)(8)(7)(6))+1/252]/(127/252)=[10/252+1/252]/(127/252)=11/127.

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Common mistakes

Condition on being above the mean before identifying the no-green cases; only RRBBB and BBBBB qualify.

Solution notes

All five asks/marks, original partially filled table, complete final table, both general score-six products and all algebra stages, integer equation, all missing-entry sequential products/order factors, exact/rounded mean and tail, and every conditional-probability stage retained.

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