anonymous
  • anonymous
A fair coin is flipped four times. What is the probability of getting heads at least once? Write your answer as a simplified fraction.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
pls use the complement approach
anonymous
  • anonymous
That's what I would have thought anyway. What's the probability of not getting heads at all?
anonymous
  • anonymous
ah the probability is getting all tails

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anonymous
  • anonymous
Which is?
anonymous
  • anonymous
it is 1/2*1/2*1/2*1/2
anonymous
  • anonymous
1/16
anonymous
  • anonymous
so whts ur reply?
anonymous
  • anonymous
Yes, that's correct. Now, there are only two possibilities: Either we get no heads or at all, or we get at least one. So, since the probabilities must add to one, what do we get?
anonymous
  • anonymous
p(of having atlesat one head)=1-p(of having no head)
anonymous
  • anonymous
1-1/16=15/16
anonymous
  • anonymous
there ya go
anonymous
  • anonymous
welll my maths sucks
anonymous
  • anonymous
Hahah well you handled that fine
anonymous
  • anonymous
is there any algorith to solve probability ques
anonymous
  • anonymous
No, they need to be solved on a case-by-case basis.
Directrix
  • Directrix
You can use Binomial Probability Distributions to solve the problem. It is an algorithm of sorts. P(at least 1 H) = 1 - P(0H) = 1 - C(4,0) (1/2)^0 (1/2)^4 = 1 - 1 (1/16) = 15/16. Note that C(4,0) is combinatorial notation for 4 choose 0. Check out this video on binomial probabilities: http://www.youtube.com/watch?v=xNLQuuvE9ug
anonymous
  • anonymous
what is a combinatoral notation?
Directrix
  • Directrix
http://mathworld.wolfram.com/Combination.html Also, http://mathworld.wolfram.com/Combinatorics.html
anonymous
  • anonymous
got it u are taking about combination notation of nCr.....
anonymous
  • anonymous
murray what is the other notations for combinations
Directrix
  • Directrix
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cristiann
  • cristiann
You should get yourself used to think in terms of "events". "the probability of getting all tails" is (1/2)^4 because the events are independent: the event of getting a tail at one throw doesn't affect the event of getting a tail at another throw....
anonymous
  • anonymous
thanx cristann ....is this rule applicable in a pack of cards?

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