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If the probability that Green Bay wins the Super Bowl is 1/12 and the

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Bunuel
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If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   04 Feb 2015, 08:56
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If the probability that Green Bay wins the Super Bowl is 1/12 and the probability that Milwaukee wins the World Series is 1/20, what is the approximate probability that either Green Bay wins the Super Bowl or Milwaukee wins the World Series?

A. 1/240
B. 1/12
C. 1/8
D. 1/7
E. 4/3

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C
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sterling19
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   04 Feb 2015, 09:26
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Winning the Super Bowl and winning the World Series are independent events; one outcome does not affect the other outcome.

\(P(A or B) = P(A) + P(B) - P(A and B)\)
\(P(A and B) = P(A) * P(B)\)

1/12 + 1/20 - (1/12 * 1/20)
= 5/60 + 3/60 - 1/120
= 8/60 - 1/120
= 16/120 - 1/120
= 15/120
= 1/8

The correct answer is C.
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chetan2u
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   04 Feb 2015, 09:35
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the two events being independent gives us P(super bowl,world cup)=not winning super bowl*winning world cup + winning super bowl*not winning world cup =19/20*1/12+1/20*11/12=30/240=1/8 ans C
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   09 Feb 2015, 05:23
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Bunuel wrote:
If the probability that Green Bay wins the Super Bowl is 1/12 and the probability that Milwaukee wins the World Series is 1/20, what is the approximate probability that either Green Bay wins the Super Bowl or Milwaukee wins the World Series?

A. 1/240
B. 1/12
C. 1/8
D. 1/7
E. 4/3

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VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: (C)

The probability of “X or Y” can be expressed as (Prob X) + (Prob Y) - (Prob of Both). In our case, this is 1/12 + 1/20 – (1/12 * 1/20), or 20/240 + 12/240 – 1/240, or 31/240, which is approximately 1/8.
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   18 Jun 2016, 06:14
This one is mathematically strange,

If you use the formula P(A or B) you get 31/240

If you use the formula P(A) * P(not B) + P(B) * P(not A) you get 1/8 !!!

How is this possible?.. Both methods are correct I think and we get a different result?... (very close but math is an exact science, right?)
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If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   02 Dec 2017, 10:34
Hi ,

As has been discussed above posts , P(AUB) = P (A)+ P (B)- P( Both A and B)

= 1/12 + 1/20 -( 1/12*1/20)
~ 1/12 + 1/20 ( since 1/ 240 will be too small and considering less time duriing exam, we can ignore that)
~ 32/240 = 2/15 = 0.13
approximately equal to 1/8 ( 0.125)

So Answer is C.
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   11 Jul 2020, 22:52
We can use conditional probability -

= P(Green Bay wins)*P(Milwaukee doesn't win) + P(Green Bay doesn't win)*P(Milwaukee wins)
= 1/12 * 19/20 + 11/12 * 1/20
=(19+11)/12*20
=1/8 (OPTION C)
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   04 Mar 2022, 12:59
Experts,

Is conditional probability tested on GMAT?
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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   04 Mar 2022, 14:38
People arriving at 1/8 as the exact answer are ignoring the outcome when BOTH occur, which is 1/240,and accounts for the exact answer of 31/240 instead of 30/240.

OR includes AnotB, BnotA and AB

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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the [#permalink] New post   07 Apr 2023, 02:55
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