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A drawer holds 4 red hats and 4 blue hats. What is the [#permalink]

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12 Apr 2006, 19:18

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A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer?
(a) 1/8
(b) Â¼
(c) Â½
(d) 3/8
(e) 7/12

A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer? (a) 1/8 (b) Â¼ (c) Â½ (d) 3/8 (e) 7/12

A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer? (a) 1/8 (b) Â¼ (c) Â½ (d) 3/8 (e) 7/12

I got 2 * ((4c3)*(4c1)) / (8c4) = 16/35. none of the answers, i might be wrong

If we use binominal formula (0,5+0,5)^4 the prob of getting 3R1B or 3B1R is 3C4*0,5*0,5^3=4/16=1/4

I thought about this approach too, even though it doesn't make too much sense, cause it would imply that every time you get one hat you put it back in the drawer and then pick a hat again, I would assume there is not replacement, but anyway even if we take BG's approach the answer would not be 1/4 but 1/2 as we have to consider 3reds - 1 blue or 1 red - 3 blues.

so taking the options from the question it would be