# Need a stats wiz… Simple question…

Six Sigma – iSixSigma Forums Old Forums General Need a stats wiz… Simple question…

Viewing 13 posts - 1 through 13 (of 13 total)
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• #46470

Anderson
Participant

How do I figure out the probablitiy of getting at least (1) heads when flipping a coin (3) times? (given that the prob. of getting a heads on one flip is 50%)
I can’t seem to remember how to do this…
Thanks,
-Mike

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#153627

anonimus
Participant

the events could be
H H H;  HHC; HCC ; CCC.   3 opportunity on 4
the probability you can get at least one Head is = .75

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#153629

Cone
Participant

This is a binomial problem and your answer is wrong because you did not account for the number of ways some thing can happen
0 heads can only happen one way and the probability is 0.125
1 head can happen three different ways (HTT,THT, or TTH) and the probability of each is 0.125 for a total of 0.375
2 heads can happen three different ways (HHT,HTH, or THH) and the probability of each is 0.375
3 heads can only happen one way and the probability is 0.125.
So the probability of 1 or more heads = 1- probability of 0 heads = 0.875.

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#153633

Bower Chiel
Participant

As with many probability probability problems this one may be solved quickly by looking at the complementary event to the one that you are interested in.P(At least one head)
=1-P(All three coins show tails)
=1-0.5*0.5*0.5
=0.875Sorry anonimus but I think you should have listed: –
HHH HHT HTH THH HTT THT TTH TTT.Bower Chiel

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#153636

jediblackbelt
Participant

Maybe I am reading his question wrong, but I read it as if I flip a coin 3 times what is the probability that only one of them is heads.  If that is the case then wouldn’t you use the binomial probability?
I come up with a .375 chance that flipping a coin 3 times that only 1 of the coin flips will be heads.
Am I off base?

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#153640

Cone
Participant

Look at the words “at least” in the original problem.

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#153643

BritW
Participant

If it matters at this point – Gary is right.  Probability of filpping at least 1 head = 1-P(flipping 0 heads) = 1-.125 = .875.
=1-{(3!/(0!(3-0)!)) x .5(to 0 power) x .5(to 3-0 power)}
=1- {(1) x (1) x .125}
= .875

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#153649

Anderson
Participant

Thanks for the replies.  I used the 1-p approach, but when I changed the probability of getting a heads to 20% (instead of 50%) my probability of getting at least one heads on three flips went up… which doesn’t make sense.
(.5 x .5 x .5) = .125 = 1-p = 1-.125 = 87.5%
(.2 x .2 x .2) = .008 = 1-.008 = 99.2%
What am I doing wrong?

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#153650

Cone
Participant

It’s 1 – probability of three tails which is 1 – .8*.8*.8 or .488.

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#153651

BTDT
Participant

Mikea:In your second calculation you should use three p(tails) and subract from unity. You used three p(heads).Cheers, BTDT

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#153652

Anderson
Participant

Doh. Ok.. I feel like an idiot… not the first time either. Thanks all. =o)

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#153663

jediblackbelt
Participant

Cool.  Not off base, just misread.  Good thing this isn’t a stats test.  Thanks for the clarification.

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#153670

goodie
Participant

MikeA,
the diagram would be:

1st flip                      H                                        T
2nd flip            H              T                          H             T
3rd flip       H       T       H       T               H      T      H     T
HHH HHT HTH HTT           THH THT  TTH  TTT
combination with at least 1 H is 7 out of 8, so
the probability of having at least 1 Head is 7/8 = .875

Goodie

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