Question

An Ad-hoc committee is to be established by Planning department to oversee open space issues in central city area. Committee members are to be selected by lot from among 10 candidates, of which 4 are pro-growth and 6 are zero-growth advocates . At issue whether the committee should have 3 or 5 members. Which size would be more likely to produce a pro-growth majority? Show your calculations!!

Answer 1

Please note ⁿC_r = n! / [(n-r)!*r!]. Also some standard formulae are as below.

(i) ⁿC_r = ⁿC_n-r eg ⁵C₂ = ⁵C₃

(ii) ⁿC₁ = n eg ⁵C₁ = 5

(iii) ⁿC₀ = 1 eg ⁵C₀ = 1 and

(iv) ⁿC_n = 1 eg ⁵C₅ = 1

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If the committee has 3 members, A majority will occur when the committee has 2 pro growth and 1 non growth member or a;; 3 are pro growth members.

Number of committees with 2 pro growth and 1 zero growth = ⁴C₂ x ⁶C₁ = 6 * 6 = 36

Number of committees with 3 pro growth and 0 zero growth = ⁴C₃ x ⁶C₀ = 4 * 1 = 4

Therefore Total favorable committees = 36 + 4 = 4

Total possible 3 member committees = ¹⁰C₃ = 120

P(committees with 2 pro growth and 1 zero growth) = 36/120 = 3/10 = 0.3

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If the committee has 5 members, A majority will occur when the committee has 3 pro growth and 2 non growth member or a committee has 4 are pro growth members and 1 zero growth member. (committee with 5 pro groth members is not possuble as there are only 4 of them)

Number of committees with 3 pro growth and 2 zero growth = ⁴C₃ x ⁶C₂ = 4 * 15 = 90

Number of committees with 4 pro growth and 1 zero growth = ⁴C₄ x ⁶C₁ = 4 * 6 = 24

Therefore Total favorable committees = 90 + 24 = 114

Total possible 3 member committees = ¹⁰C₅ = 252

P(committees with 2 pro growth and 1 zero growth) = 114/252 = 19/42 = 0.45

Therefore the probability that a 5 member committee produces a majority of pro growth is larger/more than that of a 3 member committee.

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