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#1 2006-10-01 03:09:56
- abc4616
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- Registered: 2006-10-01
- Posts: 9
Need Help
Suppose a coin is tossed until a head appears. Let p be the probability of a head on any given toss. Define random variable X to be the number of tosses required to obtain a head.
a) State the set of values that X can take on. Is this finite or countably infinite set?
b) What is the PMF (probability mass function) of X? If p = 1/4, what is the probability that X=10?
Can any one help on this question??
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#2 2006-10-01 04:24:40
- polylog
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- Registered: 2006-09-28
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Re: Need Help
I think part (a) can be approached like this:
The values that X can take on are the possible number of times that we have to toss the coin until a head appears.
We might have to toss it once, or twice, or three times.. ie, X belongs to the set:
{1, 2, 3, 4, 5, 6, 7, 8, .... }
This is just the set of natural numbers, N, which is countably infinite.
(If I'm wrong surely someone will correct me 
)
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#3 2006-10-01 06:08:48
- Ricky
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Re: Need Help
Sounds right to me, polylog.
For b, are you allowed to use the binomal distribution?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#4 2006-10-01 06:39:10
- Patrick
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- Registered: 2006-02-24
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Re: Need Help
I'm not sure about b), but polylogs' answer to a) seems correct
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#5 2006-10-01 07:11:04
- abc4616
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Re: Need Help
For b, you can use binomial distribution!!
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#6 2006-10-01 07:27:07
- mathsyperson
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- Registered: 2005-06-22
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Re: Need Help
I'm not sure that the binomial distribution is what we need here.
In that, you have a fixed number of trials but here you just keep going until you get a head.
I think the answer you want is P(X=x) = p * (1-p)^(x-1).
So, if p was 1/4, then P(X=10) would be 1/4 * (3/4)^9, which is around 0.0188.
Why did the vector cross the road?
It wanted to be normal.
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#7 2006-10-01 08:24:24
- Ricky
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Re: Need Help
What you need is the negative binomial distribution:
The probability of having the xth success land on the nth trial, where p = probability of a single sucess
P(X) = ((n-1) choose (x-1)) * (p)^x * (1 - p)^(n-x)
That comes out to be:
P(1) = (10-1) choose (1-1) * (1/4)^1 * (1 - 1/4)^(10-1) = 0.018771
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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