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

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If the probability that Green Bay wins the Super Bowl is 1/12 and the  [#permalink]

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New post 04 Feb 2015, 08:56
2
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A
B
C
D
E

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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

Kudos for a correct solution.

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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the  [#permalink]

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New post 04 Feb 2015, 09:26
1
3
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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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the  [#permalink]

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New post 04 Feb 2015, 09:35
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]

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New post 09 Feb 2015, 05:23
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

Kudos for a correct solution.


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If the probability that Green Bay wins the Super Bowl is 1/12 and the  [#permalink]

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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]

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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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If the probability that Green Bay wins the Super Bowl is 1/12 and the &nbs [#permalink] 02 Dec 2017, 10:34
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