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island-boy
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I got really really confused by this supposedly easy discrete probability problem:
The problem asks:
a)toss a die until a "6" appears. Find the probability distribution of X where X is the number of tosses neded to obtain the first six.
b) Prove that the summation of P(x) from x = 1 to n (not infinity!) is equal to 1.
c) Determine P(X = 2k + 1) for every k an element of a natural number.
d) Find the distribution function of X in c)
Here's what I got:
for a) obviously, the answer is a geometric distribution as the answer is:
P(X=x) = f(x) = (5/6)^(x-1) * (1/6) for x = 0,1,2,3,... and 0 otherwise
for c) I just substitute x with 2k + 1, thus,
P(X = 2k + 1) = (5/6)^(2k) * (1/6) for any k, natural number
what I can't understand is b and d.
How the heck are you suppose to prove b?
since P(x) is a geometric series, then its summation from 1 to n is given by a(1 -r^n)/(1-r) which is equal to 1 - (5/6)^n when a = 1/6 and r = 5/6...this summation is equal only to 1 when n reaches infinity! I asked the TA who gave this question about this, and he keeps insisting it should be up to n, not up to infinity. (if it is up to infinity, the summation is just a/1-r = 1!). Am I just crazy and the TA is right, or can it be proven that the summation up to n is equal to 1?
And for d), is it possible to get the distibution function for X = 2k+1? not the probability (which is c), but the distribution itself. I asked if this is the conditional distribution f(x| x = 2k +1), and the TA says it isn't. It's the distribution itself. Is this even possible? Again, I hope I'm not crazy.
This has been driving me bananas for the last 2 days...
any help is appreciated. thanks
The problem asks:
a)toss a die until a "6" appears. Find the probability distribution of X where X is the number of tosses neded to obtain the first six.
b) Prove that the summation of P(x) from x = 1 to n (not infinity!) is equal to 1.
c) Determine P(X = 2k + 1) for every k an element of a natural number.
d) Find the distribution function of X in c)
Here's what I got:
for a) obviously, the answer is a geometric distribution as the answer is:
P(X=x) = f(x) = (5/6)^(x-1) * (1/6) for x = 0,1,2,3,... and 0 otherwise
for c) I just substitute x with 2k + 1, thus,
P(X = 2k + 1) = (5/6)^(2k) * (1/6) for any k, natural number
what I can't understand is b and d.
How the heck are you suppose to prove b?
since P(x) is a geometric series, then its summation from 1 to n is given by a(1 -r^n)/(1-r) which is equal to 1 - (5/6)^n when a = 1/6 and r = 5/6...this summation is equal only to 1 when n reaches infinity! I asked the TA who gave this question about this, and he keeps insisting it should be up to n, not up to infinity. (if it is up to infinity, the summation is just a/1-r = 1!). Am I just crazy and the TA is right, or can it be proven that the summation up to n is equal to 1?
And for d), is it possible to get the distibution function for X = 2k+1? not the probability (which is c), but the distribution itself. I asked if this is the conditional distribution f(x| x = 2k +1), and the TA says it isn't. It's the distribution itself. Is this even possible? Again, I hope I'm not crazy.
This has been driving me bananas for the last 2 days...
any help is appreciated. thanks
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