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Three friends, Andy, Bob, and Chad enter a quiz competition.
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06 Feb 2024, 01:37
06 Feb 2024, 01:37
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Question Stats:
33% (01:05) correct
66% (02:12) wrong based on 3 sessions
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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. \(\dfrac{3(a-1)(a-2)}{a^3}\)
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
07 Feb 2024, 10:38
how B & C
Re: Three friends, Andy, Bob, and Chad enter a quiz competition.
[#permalink]
08 Feb 2024, 00:18
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
