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Date: 02/12/2002 at 09:11:18
From: Courtney
Subject: Probability of getting an even number of heads when tossing a
biased coin

In my book, it says that the probability of getting an even number of
heads when a coin is tossed N times, when the probablity of getting a
head is p and tail is q = 1-p, is .5(1+(q-p)^N).

Can you explain how they got this?

Thanks,
Courtney
```

```
Date: 02/12/2002 at 19:38:20
From: Doctor Anthony
Subject: Re: Probability of getting an even number of heads when
tossing a biased coin

Let u(r) be the probability that there is an even number of heads
after r throws. So 1-u(r) is the probability of an odd number of heads
after r throws.

The probability of an even number of heads after r throws is q times
the probability of an even number of heads after r-1 throws plus p
times the probability of an odd number of heads after r-1 throws.

So    u(r) = q.u(r-1) + p.[1-u(r-1)]

u(r) = u(r-1).(q-p) + p

= [u(r-2).(q-p) + p](q-p) + p

= u(r-2).(q-p)^2 + p(q-p) + p

= [u(r-3).(q-p) + p](q-p)^2 + p(q-p) + p

= u(r-3).(q-p)^3 + p(q-p)^2 + p(q-p) + p

and so on down to

u(0).(q-p)^r + p[(q-p)^(r-1) + (q-p)^(r-2) + ...+ 1]

1 - (q-p)^r
u(0).(q-p)^r + p[-------------]
1 - (q-p)

1 - (q-p)^r
u(0).(q-p)^r + p[-------------]
p+q - q + p

u(0).(q-p)^r + (1/2)[1 - (q-p)^r]

and u(0) = 1 so this reduces to

(q-p)^r + 1/2 - (1/2).(q-p)^r

=  1/2 + (1/2)(q-p)^r

=  (1/2)[1 + (q-p)^r]    as required

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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