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by bblast » Mon Jun 20, 2011 8:35 am
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

9%
12%
16%
25%
20%

[spoiler]i guessed : 1/4 *2/4 * 3/4 * 4/4 = 3/32 = 9% , which is the OA :twisted: [/spoiler]

kindly explain
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by Frankenstein » Mon Jun 20, 2011 8:41 am
Hi,
Each of the player has 4 choices.
So, total number of combinations is 4^4
Four people can select 4 different numbers is 4! ways
So, probability is 4!/4^4 = 3/32
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by GMATGuruNY » Mon Jun 20, 2011 2:07 pm
bblast wrote:As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

9%
12%
16%
25%
20%

[spoiler]i guessed : 1/4 *2/4 * 3/4 * 4/4 = 3/32 = 9% , which is the OA :twisted: [/spoiler]

kindly explain
The first person can choose any number he wants.

P(second person chooses a number not yet chosen) = 3/4. (Out of the 4 numbers, 3 haven't been chosen yet).
P(third person chooses a number not net chosen) = 2/4. (Out of the 4 numbers, 2 haven't been chosen yet.)
P(fourth person chooses a number not yet chosen) = 1/4. (Out of the 4 numbers, 1 hasn't been chosen yet.)

Since we want all of the events above to happen together, we multiply the fractions:

3/4 * 2/4 * 1/4 = 3/32 = 9/96 ≈ 9%.

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