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Bunuel
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 11 Jun 2019, 01:41
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81% (01:45) correct 19% (02:13) wrong based on 286 sessions

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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?


A. \(\frac{8}{125}\)

B. \(\frac{36}{125}\)

C. \(\frac{108}{125}\)

D. \(\frac{117}{125}\)

E. \(\frac{124}{125}\)
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D
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 11 Jun 2019, 02:27
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Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?


A. \(\frac{8}{125}\)

B. \(\frac{36}{125}\)

C. \(\frac{108}{125}\)

D. \(\frac{117}{125}\)

E. \(\frac{124}{125}\)
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Prob=1-All three are green; 1-(2/5)^3=117/125 IMO D
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 11 Jun 2019, 05:51
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Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?


A. \(\frac{8}{125}\)

B. \(\frac{36}{125}\)

C. \(\frac{108}{125}\)

D. \(\frac{117}{125}\)

E. \(\frac{124}{125}\)
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Probability not all disks are green = 1 - Probability that all disks chosen are green

--> Prob that all disks chosen are green = 2/5*2/5*2/5 = 8/125
--> Required Probability = 1 - 8/125 = 117/125

IMO Option D

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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 11 Jun 2019, 11:39
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Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?


A. \(\frac{8}{125}\)

B. \(\frac{36}{125}\)

C. \(\frac{108}{125}\)

D. \(\frac{117}{125}\)

E. \(\frac{124}{125}\)
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all are green ; 8/125
and not green 1-8/125 ; 117/125
IMO D
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 12 Jun 2019, 02:53
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Solution

Given:
    • Total number of bags = 3
    • Each bag has
      o 2 green disks
      o 3 blue disks

To Find:
    • The probability that not all of the disks chosen are green, if one disk is chosen at random from each of the bags

Approach & Working Out:
    • The probability that not all of the disks chosen are green = 1 – the probability that all the three disks chosen are green = \(1 – (\frac{2}{5}) * (\frac{2}{5}) * (\frac{2}{5}) = 1 – \frac{8}{125} = \frac{117}{125}\)

Hence, the correct answer is Option D
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 17 Jun 2019, 17:53
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Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?


A. \(\frac{8}{125}\)

B. \(\frac{36}{125}\)

C. \(\frac{108}{125}\)

D. \(\frac{117}{125}\)

E. \(\frac{124}{125}\)
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We can use the formula:

P(not all of the disks chosen are green) = 1 - P(all of the disks chosen are green)

Since P(all of the disks chosen are green) = 2/5 x 2/5 x 2/5 = 8/125, then P(not all of the disks chosen are green) = 1 - 8/125 = 117/125.

Answer: D
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 01 Jul 2020, 03:51
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Bunuel
In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing else. If one disk is chosen at random from each of the bags, what is the probability that not all of the disks chosen are green?


A. \(\frac{8}{125}\)

B. \(\frac{36}{125}\)

C. \(\frac{108}{125}\)

D. \(\frac{117}{125}\)

E. \(\frac{124}{125}\)
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P(G)=2/5 from each bag
P(G) from all bags=8/125
P(Not all G)=1-8/125=117/125
=117/125
Or we can use the time consuming method:)
Blue, Blue, Blue- 27/125
Blue, Blue, Green- 3/5*3/5*2/5*(!3/!2)-54/125
Blue, Green, Green-3/5*2/5*2/5*(!3/!2)-36/125
Total=27+54+36/125
=117/125
D:)
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In each of 3 bags, there are 2 green disks, 3 blue disks, and nothing 22 Jul 2025, 16:06
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