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Three friends, Andy, Bob, and Chad enter a quiz competition.
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06 Feb 2024, 01:37

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Three friends, Andy, Bob, and Chad enter a quiz competition. The probability of Bob winning the competition is double that of Andy's and the probability of Chad winning the competition is thrice that of Andy's. If the probability of Andy winning the competition is a, which of the following could be the probability of winning the competition by any of the three friends?

Indicate all correct options.

A. \(\dfrac{(a-2)(a-3)}{a^3}\)

B. \(\dfrac{2(a-1)(a-3)}{a^3}\)

C. \(\dfrac{3(a-1)(a-2)}{a^3}\)

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Indicate all correct options.

A. \(\dfrac{(a-2)(a-3)}{a^3}\)

B. \(\dfrac{2(a-1)(a-3)}{a^3}\)

C. \(\dfrac{3(a-1)(a-2)}{a^3}\)

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Re: Three friends, Andy, Bob, and Chad enter a quiz competition.
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07 Feb 2024, 10:38

how B & C

Re: Three friends, Andy, Bob, and Chad enter a quiz competition.
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08 Feb 2024, 00:18

Expert Reply

Andy winning is 1/a

Bob winning is 1/a*2=2/a

Chad is three times 1/a*3=3/a

Andy NOT winning is 1-1/a=a-1/a

Bob winning is 1-2/a=a-2/a

Chad winning is 1-3/a=a-3/a

1) Andy wins and the others lose is

1/a*a-2/a=(a-2)(a-3)/a^3 Correct

2) Bob wins and the others lose

a-1/a*2/a*a-3/a=2(a-1)(a-3)/a^3 Correct

3) Chad wins and the others lose

a-1/a*a-2/a-3/a= 3(a-1)(a-2)/a^3 Correct

_________________

Bob winning is 1/a*2=2/a

Chad is three times 1/a*3=3/a

Andy NOT winning is 1-1/a=a-1/a

Bob winning is 1-2/a=a-2/a

Chad winning is 1-3/a=a-3/a

1) Andy wins and the others lose is

1/a*a-2/a=(a-2)(a-3)/a^3 Correct

2) Bob wins and the others lose

a-1/a*2/a*a-3/a=2(a-1)(a-3)/a^3 Correct

3) Chad wins and the others lose

a-1/a*a-2/a-3/a= 3(a-1)(a-2)/a^3 Correct

_________________

Re: Three friends, Andy, Bob, and Chad enter a quiz competition.
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17 Mar 2024, 20:17

Carcass wrote:

Andy winning is 1/a

Bob winning is 1/a*2=2/a

Chad is three times 1/a*3=3/a

Andy NOT winning is 1-1/a=a-1/a

Bob winning is 1-2/a=a-2/a

Chad winning is 1-3/a=a-3/a

1) Andy wins and the others lose is

1/a*a-2/a=(a-2)(a-3)/a^3 Correct

2) Bob wins and the others lose

a-1/a*2/a*a-3/a=2(a-1)(a-3)/a^3 Correct

3) Chad wins and the others lose

a-1/a*a-2/a-3/a= 3(a-1)(a-2)/a^3 Correct

Bob winning is 1/a*2=2/a

Chad is three times 1/a*3=3/a

Andy NOT winning is 1-1/a=a-1/a

Bob winning is 1-2/a=a-2/a

Chad winning is 1-3/a=a-3/a

1) Andy wins and the others lose is

1/a*a-2/a=(a-2)(a-3)/a^3 Correct

2) Bob wins and the others lose

a-1/a*2/a*a-3/a=2(a-1)(a-3)/a^3 Correct

3) Chad wins and the others lose

a-1/a*a-2/a-3/a= 3(a-1)(a-2)/a^3 Correct

How is the probability 1/a? Could you please explain?

Re: Three friends, Andy, Bob, and Chad enter a quiz competition.
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18 Mar 2024, 02:09

1

Expert Reply

The probability is always at most 1

So if your probability to win is a this a is <1. For example my probability to win is 25%. Therefore, my probability is 1/4=25%

More on probability is here https://gre.myprepclub.com/forum/gre-pr ... tml#p83091

_________________

So if your probability to win is a this a is <1. For example my probability to win is 25%. Therefore, my probability is 1/4=25%

More on probability is here https://gre.myprepclub.com/forum/gre-pr ... tml#p83091

_________________

gmatclubot

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