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

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Manager
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This one has been done before : A coin is flipped 5 [#permalink]  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 ?
Eternal Intern
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Tzolkin [#permalink]  03 Aug 2003, 08:01
2^5 = 32 outcome

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

am I right
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Ride em cowboy

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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
Eternal Intern
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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.

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Ride em cowboy

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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.
Eternal Intern
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VT [#permalink]  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. Spend more time with algebra and geometry, why are we all obsessed about probability
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Ride em cowboy

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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

Manager
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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
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