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Each of the four children has a bag with five different colored discs in it

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by BTGModeratorVI » Fri Apr 10, 2020 8:15 am

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Answers

A

B

C

D

E

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Difficulty—

Each of the four children has a bag with five different colored discs in it; red blue green yellow and orange. If each child in succession randomly chooses exactly one disc from his bag to flip into a hat. What is the probability that hat will no hold no repeated colours at the end flipping?


A) 1/5
B) 6/625
C) 24/625
D) 24/125
E) 6/125

Answer: D
Source: 800Score
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Re: Each of the four children has a bag with five different colored discs in it

by Brent@GMATPrepNow » Sat Apr 11, 2020 6:25 am
BTGModeratorVI wrote: ↑
Fri Apr 10, 2020 8:15 am
 Each of the four children has a bag with five different colored discs in it; red blue green yellow and orange. If each child in succession randomly chooses exactly one disc from his bag to flip into a hat. What is the probability that hat will no hold no repeated colours at the end flipping?


A) 1/5
B) 6/625
C) 24/625
D) 24/125
E) 6/125

Answer: D
Source: 800Score
As with most probability questions on the GMAT, we can use either counting techniques or probability rules.
Let's use probability rules:

P(no repeated colors) = P(ANY color disc is chosen 1st AND color of 2nd disc is different from 1st disc AND color of 3rd disc is different from 1st and 2nd disc AND color of 4th disc is different from other discs)
= P(ANY color disc is chosen 1st) x P(color of 2nd disc is different from 1st disc) x P(color of 3rd disc is different from 1st and 2nd disc) x P(color of 4th disc is different from other discs)
= 1 x 4/5 x 3/5 x 2/5
= 24/125

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Re: Each of the four children has a bag with five different colored discs in it

by deloitte247 » Sun Apr 12, 2020 4:31 am
There are 4 children with bags that holds 5 different colors, probability of the first child choosing a unique disc = 5/5; color disc remains 4

Probability of second child choosing a unique disc from the remaining 4 = 4/5; color disc remains 3

Probability of third child choosing a unique disc from the remaining 3 = 3/5; color disc remains 2

Probability of fourth child choosing a unique disc from the remaining 2 = 2/5; color disc remains 1

Probability of the fifth child choosing the remaining unique disc = 1/5

Probability of all 5 children choosing unique color disc without repeated colors
= 5/5 * 4/5 * 3/5 * 2/5 * 1/5 = 24/125

Answer = D
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