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pls use the complement approach
That's what I would have thought anyway. What's the probability of not getting heads at all?
ah the probability is getting all tails
it is 1/2*1/2*1/2*1/2
so whts ur reply?
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?
p(of having atlesat one head)=1-p(of having no head)
there ya go
welll my maths sucks
Hahah well you handled that fine
is there any algorith to solve probability ques
No, they need to be solved on a case-by-case basis.
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
what is a combinatoral notation?
http://mathworld.wolfram.com/Combination.html Also, http://mathworld.wolfram.com/Combinatorics.html
got it u are taking about combination notation of nCr.....
murray what is the other notations for combinations
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....
thanx cristann ....is this rule applicable in a pack of cards?