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# A fair coin is thrown 5 times. What is the probability of

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A fair coin is thrown 5 times. What is the probability of [#permalink]

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03 Oct 2006, 09:41
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A fair coin is thrown 5 times. What is the probability of throwing at least 2 heads?

What is the best method to solve such question?
SVP
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03 Oct 2006, 11:03
I prefer to search the opposite probability.

o Probability of "no head" = (1/2)^5
o Probability of "1 head" = 5/(2)^5 (5 possible positions)

Thus, probability of "at least 2 heads" = 1 - 1/16 - 1/32 = 29/32
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03 Oct 2006, 12:22
davvy75 wrote:
A fair coin is thrown 5 times. What is the probability of throwing at least 2 heads?

What is the best method to solve such question?

Use binomial theorem to solve this.

P0 = 5C0 X (1/2)^0 X (1/2) ^ 5 = (1/2)^5
P1 = 5C1 X (1/2)^1 x (1/2) ^ 4 = 5. (1/2)^5

Your answer should be 1 - P0 - P1 = 1 - 6/32 = 26/32

Alternatively you could find out P2, P3, P4 and P5 and add them up
P2 + P3 + P4 + P5 = (1/2)^5 X (5C2 + 5C3 + 5C4 + 5C5) = 1/32 (10+10+5+1) = 26/32

which can be reduced to 13/16. As you can see, the 1st method is shorter. I hope this is the answer
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03 Oct 2006, 12:34
5*(1/2)^2 = 5/32 not 1/16

Futuristic, of course, is right :D
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04 Oct 2006, 20:31
P(atleast 2 heads) = 1- P(more than 3 tails)

P(4 tails) = 1/2^4 = 1/16

P(5 tails) = 1/2^5 = 1/32

So, P (atleast 2 heads) = 29/32
I am wrong, everybody else is right!

aaaarghhh!!! Probability and permutations and combinations are going to be the death of me!
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05 Oct 2006, 00:23
It must be 13/16.

Prob of getting atleast two heads = 1- Probability of getting atmost 1 head
ie 1- [prob of getting 0 heads (i.e all tails) + prob of getting 1 head]
ie 1- (5C0/2^5 +5C1/2^5)
ie 1- 6/32 = 26/32=13/16.

The other way is
i.e (5C2+5C3+5C4+5C5)/2^5
i.e 26/32= 13/16
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