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This one has been done before : A coin is flipped 5

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This one has been done before : A coin is flipped 5 [#permalink] New post 02 Aug 2003, 16:15
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This one has been done before :

A coin is flipped 5 times. What is the probability that there are at least 3 heads ?
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Tzolkin [#permalink] New post 03 Aug 2003, 08:01
2^5 = 32 outcome

2C1 + 2C1 + 2C1
------------------
32

am I right :?:
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 [#permalink] New post 03 Aug 2003, 09:54
10C3+5C1+1C1
---------------------=16/32=1/2
32

Or

at least 3 means 3, 4, and 5 heads
the opposite cases are 0, 1, and 2 heads
the rule of symmetry gives 3/6=1/2
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 [#permalink] New post 03 Aug 2003, 10:24
coin is flipped 5 times. What is the probability that there are at least 3 heads ?

Stolyar, do you realize 10C3 works out to 120, I'm lost here.

:madd
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 [#permalink] New post 03 Aug 2003, 12:04
Can you not use the Binomial Equation ?

P = 5C3*(1/2)^3*(1/2)^2

Last edited by tzolkin on 03 Aug 2003, 15:04, edited 1 time in total.
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VT [#permalink] New post 03 Aug 2003, 12:51
Although I've never heard of it until now, Tzolkin, if you state we should spend more time doing "ETS" style, then don't worry about this question. :wink: Spend more time with algebra and geometry, why are we all obsessed about probability :roll:
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 [#permalink] New post 04 Aug 2003, 22:28
I agree that Binomial method can be applied to this problem

p1 = probability that it is Heads = 1/2
p2 = probablity that it is not Heads = 1/2

P = 5C3 (p1)^3 (p2)^2 = 0.3215

some one please confirm the answer.
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 [#permalink] New post 04 Aug 2003, 23:29
tzolkin wrote:
Can you not use the Binomial Equation ?

P = 5C3*(1/2)^3*(1/2)^2


Tzolkin

The case you state here is only for 3 heads. The question asks for ATLEAST three heads = 3 or 4 or 5 heads.

Therefore, using bionomial equation

P(atleast 3 heads) = 5C3+5C4+5C5 / (1/2)^5 = 16/32 = 1/2
  [#permalink] 04 Aug 2003, 23:29
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This one has been done before : A coin is flipped 5

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