The difficulty of this question lays in its ambiguity. Thank you to the community for clearing up what exactly this question is asking for. Once the fog is gone, its very simple.
BTW I went wrongly with A, because I couldn't tell how many boxes could be open, or if all boxes could be open, so I made my own (wrong) conclusion. Once the rules were laid out however, it was simple to follow and answer.
Game show
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Atilla the Hulk
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nikhilgmat31
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Why not to consider other possibilities of greater than 1000
more than 1000 is possible in 3 ways - 1001 or 1100 or 1101
P(1000) = 1/3
P(1) = 1/2
P(1001) is 1/3 * 1/2 = 1/6
OR
P(1100)
P(1000) = 1/3
P(100) = 1/2
P(1100) is 1/3 * 1/2 = 1/6
OR
P(1101)
P(1000) = 1/3
P(100) = 1/2
P(1) = 1
P(1101) is 1/3 * 1/2*1 = 1/6
Total - 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Please suggest.
more than 1000 is possible in 3 ways - 1001 or 1100 or 1101
P(1000) = 1/3
P(1) = 1/2
P(1001) is 1/3 * 1/2 = 1/6
OR
P(1100)
P(1000) = 1/3
P(100) = 1/2
P(1100) is 1/3 * 1/2 = 1/6
OR
P(1101)
P(1000) = 1/3
P(100) = 1/2
P(1) = 1
P(1101) is 1/3 * 1/2*1 = 1/6
Total - 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Please suggest.
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nikhilgmat31
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- Posts: 518
- Joined: Tue May 12, 2015 8:25 pm
- Thanked: 10 times
GMATGuruNY wrote:Since I'm not associated with Veritas, I've hidden the explanation below. It will be revealed if you use the quote function to respond.DanaJ wrote:Source: Veritas Prep
On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?
(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2
Experts: Only Veritas Prep experts, please!
[spoiler]There is only 1 way to win more than $1000: the contestant must assign all of the keys correctly. To win more than $1000, he must assign the $1000 key and at least 1 other key correctly. But it's impossible to assign exactly 2 keys correctly. Once 2 keys have been assigned correctly, the 3rd key must also be assigned correctly, because there will only 1 key and 1 box left, and this 1 remaining key and the 1 remaining box will have to match.
P($1000 box gets the correct key) = 1/3 (out of the 3 keys, only 1 is correct)
P($100 box gets the correct key) = 1/2 (out of the 2 remaining keys, only 1 is correct)
P($1 box gets the correct key) = 1/1 (1 key left, and it must be correct, since the incorrect keys have already been assigned to the other 2 boxes)
Since we need all of these events to happen, we multiply the fractions:
1/3 * 1/2 * 1/1 = 1/6.
The correct answer is C.[/spoiler]
Why not to consider other possibilities of greater than 1000
more than 1000 is possible in 3 ways - 1001 or 1100 or 1101
P(1000) = 1/3
P(1) = 1/2
P(1001) is 1/3 * 1/2 = 1/6
OR
P(1100)
P(1000) = 1/3
P(100) = 1/2
P(1100) is 1/3 * 1/2 = 1/6
OR
P(1101)
P(1000) = 1/3
P(100) = 1/2
P(1) = 1
P(1101) is 1/3 * 1/2*1 = 1/6
Total - 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Please suggest.
