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A drawer holds 4 red hats and 4 blue hats. What is the

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A drawer holds 4 red hats and 4 blue hats. What is the [#permalink] New post 05 Jan 2006, 07:57
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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 and returning each hat before taking out the next one?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12
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 [#permalink] New post 05 Jan 2006, 09:03
1/2

p(drawing a red hat) = p(drawing a blue hat) = 4/8 = 1/2

exactly 3 R in 4 draw:
4 ways
BRRR
RBRR
RRBR
RRRB

each of these have prob : 1/2 * 1/2 * 1/2 * 1/2 = 1/16

thus p(drawing exactly 3 R hats) = 4 * 1/ 16 = 1/4

similarly, p(drawing exactly 3B hats) = 1/4

p(3 R or 3 B) = 1/4 + 1/4 = 1/2
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 [#permalink] New post 05 Jan 2006, 16:20
1/2 using Binomial DIST. method

4C3*(1/2)^3*(1/2) * 2 = 1/2
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 [#permalink] New post 05 Jan 2006, 18:25
P(red) = P(blue) = 4/8 = 1/2

# of ways to get 3 red hats from 4 picks = 4!/3!1! = 4 (Arrange the word RRRB)
Probability of 1 such way = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
Probability of 4 such ways = 1/16 * 4 = 1/4

The same probaility applies to blue hats.

Total = 1/4 * 2 = 1/2
  [#permalink] 05 Jan 2006, 18:25
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