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**markosheehan****Member**- Registered: 2016-06-15
- Posts: 51

the probability of getting a head on flipping a biased coin is p. the coin is flipped n times producing a sequence containing m heads and (n-m) tails what is the probability of obtaining this sequence from n flips.

i cant understand the wording

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,012

hi markosheehan

Normally you might expect the probability of flipping a head to be 1/2. If it is biased, that means the coin doesn't behave like this, but, rather, has a probability of a head that is not 1/2. eg it might be P(head) = 0.75 and P(tail) = 0.25.

We are told to use 'p' for this probability.

For each head the probability is p, so for m heads we need p^m and (1-p)^(n-m) for all the tails. In addition, there are many sequences that give these results. eg. If n = 5 and m = 2 then we could get HHTTT or HTHTT or HTTHT or HTTTH or THHTT or THTHT or THTTH or TTHHT or TTHTH or TTTHH. That's ten ways altogether. You can get this number without writing out all the possibilities by calculating 5C2 = 5!/(2!.3!) = (5 x 4)/(2 x 1) = 10.

You'll find some analysis of this here: http://www.mathsisfun.com/data/binomial … ution.html but you'll have to use P(head) = p not 0.5

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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