Question

6. There are 16 cleaning sponges in a bin at Walmart. 7 are green and the remaining 9 are blue. Suppose I randomly select 4 sponges. What is the probability that:   (Round to 4 (FOUR) decimal places.)

           

            a.         All four are green?

            b.         All four are blue?

            c.         2 or 3 are blue?

           

            d.         None are green?

            e.         3 or fewer are green?

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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Probability = Favourable Outcomes/Total Outcomes

Total Outcomes = choosing any 4 balls out of the 16 = ¹⁶C₄ = 16!/[(16-4)!*4!] = 16!/(12!*4!)

= 16 * 15 * 14 * 13 * 12! / (12! * 4 * 3 * 2 * 1) = 1820

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(a) All 4 are green: Favourable outcomes, choose 4 green balls out of 7 = ⁷C₄

= 7 * 6 * 5 * 4/ 4 * 3 * 2 * 1 = 35

Therefore the required Probability = 35/1820 = 0.0192

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(b) All 4 are blue: Favourable outcomes, choose 4 blue balls out of 9 = ⁹C₄

= 9 * 8 * 7 * 6/ 4 * 3 * 2 * 1 = 126

Therefore the required Probability = 126/1820 = 0.0692

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(c) 2 or 3 are blue = 2 blue and 2 green + 3 blue and 1 green

2 blue and 2 green = ⁹C₂ * ⁷C₂ = (9*8/2) * (7*6/2) = 36 * 21 = 756

3 blue and 1 green = ⁹C₃ * ⁷C₁ = (9*8*7/3*2) * (7) = 84 * 7 = 588

Therefore Total Favourable outcomes = 756 + 588 = 1344

Therefore the required Probability = 1344/1820 = 0.7385

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(d) None are Green = All 4 are blue: Favourable outcomes, choose 4 blue balls out of 9 = ⁹C₄

= 9 * 8 * 7 * 6/ 4 * 3 * 2 * 1 = 126

Therefore the required Probability = 126/1820 = 0.0692

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(e) 3 or fewer are green = 3 green and 1 blue + 2 green and 2 blue + 1 green and 3 blue + 0 green and 4 blue

3 green and 1 blue = ⁷C₃ * ⁹C₁ = (7*6*5/3*2*1) * (9) = 35 * 9 = 315

2 green and 2 blue = ⁷C₂ * ⁹C₂ = (7*6/2) * (9*8/2) = 21 * 36 = 756

1 green and 3 blue = ⁷C₁ * ⁹C₃ = (7) * (9*8*7/3*2) = 7 * 84 = 588

0 green and 4 blue = ⁷C₀ * ⁹C₄ = (1) * (9*8*7*6/4*3*2) = 1 * 126 = 126

Therefore the total favourable outcomes = 315 + 756 + 588 + 126 = 1785

Therefore the required Probability = 1785/1820 = 0.9808

Alternative Method: (We can also understand the above as 3 or fewer are green = 1 - greater than 3 are green

= 1 - (all 4 green and 0 blue) = ⁷C₄ * ⁹C₀ = (7*6*5*4/4*3*2*1) * (1) = 35 * 1 = 35

Therefore the required probability = 1 - (35/1820) = (1820 - 35)/1820 = 1785/1820 = 0.9808)

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