- #1
playboy
A Question Reads: "Suppose that the random Variable X is the number of failures before the first success in a series of independent Bernoulli trials with success probability p"
a) derive the probability mass function of X
b) what is the probability that X < x where x is a positive integer?
My Answers:
a) this is fairly straight forward. Its just a geometric sequence. p(x) = p(1-p)^x x = 0,1,2...
b) I AM STUCK ON THIS. I know that X < x is just the cumulative distribution function for this geometric sequence, but that just does not work well with this. I tried something different:
"the number of failures until the first sucess? P(X<x) for a geometric sequnce? that means x-1 failures in n trials? (or we could think of it as x-1 sucess in n trials if we look at a failure as a success.)
then would it be X~binomial(n,p)"
Or perhaps it means P(X=0) + P (X=1) + P(X=2) + P(X=3)... + P(X=x-1) for a geometric sequence?
Can somebody help me on it please? thanks!
a) derive the probability mass function of X
b) what is the probability that X < x where x is a positive integer?
My Answers:
a) this is fairly straight forward. Its just a geometric sequence. p(x) = p(1-p)^x x = 0,1,2...
b) I AM STUCK ON THIS. I know that X < x is just the cumulative distribution function for this geometric sequence, but that just does not work well with this. I tried something different:
"the number of failures until the first sucess? P(X<x) for a geometric sequnce? that means x-1 failures in n trials? (or we could think of it as x-1 sucess in n trials if we look at a failure as a success.)
then would it be X~binomial(n,p)"
Or perhaps it means P(X=0) + P (X=1) + P(X=2) + P(X=3)... + P(X=x-1) for a geometric sequence?
Can somebody help me on it please? thanks!
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