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by cans » Mon May 30, 2011 3:36 am
A drawer holds 4 red hats and 4 blue hats. What is the probability of having exactly 3 red hats or exactly 3 blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next one?
a)1/8
b)1/4
c)1/2
d)3/8
e)7/12
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by Anurag@Gurome » Mon May 30, 2011 3:50 am
cans wrote:A drawer holds 4 red hats and 4 blue hats. What is the probability of having exactly 3 red hats or exactly 3 blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next one?
a)1/8
b)1/4
c)1/2
d)3/8
e)7/12
Total no. of ways of taking out 4 hats randomly and replacing them = 2 * 2 * 2 * 2 = 16
Now the possible combinations are: (RRRB), (RBRRR), (BRRR), (RRBR), (BBBR), (RBBB), (BRBB), (BBRB)
There are 8 possible combinations.
Therefore, required probability = 8/16 = [spoiler]1/2[/spoiler]

The correct answer is C.
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by SoCan » Mon May 30, 2011 6:28 am
If you're familiar with Pascal's Triangle, I think that's the quickest way to do questions like this, and it makes sure you haven't accidentally forgotten a possible combination or miscounted them.

The 2^4 line of Pascal's Triange is 1 4 6 4 1. You can see right away there are (4+4) combinations with exactly three of one of the hats (1 combination of 4 blue, 1 combination of 4 red, and 6 combinations of 2 of each).

If you're not familiar with the triangle, or have forgotten it from your school days, it may be worth refreshing if you have the time. When there aren't that many iterations, it's easy enough to list out possible cases, but it gets trickier, more time consuming, and more prone to error the more the higher the number of events.
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by cans » Mon May 30, 2011 11:51 am
OA C
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