In: Math
Consider an experiment where you toss a coin as often as necessary to turn up one head.Suppose that the probability of having a tail is p(obviously probability of a head is 1−p).Assume independence between tosses.a) State the sample space.
b) Let X be the number of tosses needed to get one head. What is the support (possible values ofX)?
c) FindP(X= 1),P(X= 2) andP(X= 3).
d) Deduce the pmf of X from part c).
a)
Let H denote head and T denote tail.
We need to find the sample space for the experiment where we continuously toss the coin until we get a head. Before that first head comes, we need to have only tails as our outcome.
Now, we could get a head on the first toss in which case the event will be : H
or, we could get a head on the second toss in which case the event will be : TH
or, we could get a head on the third toss in which case the event will be : TTH
or, we could get a head on the fourth toss in which case the event will be : TTTH
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Thus, the sample space S = {H,TH,TTH,TTTH,TTTTH,....}
b)
We are given that X denotes the number of tosses required to get first head.
Now, we could use the information in part a) to help us in this part.
We could get the first head on the first toss in which case X = 1 (Corresponding to the event H in part a))
or, we could get the first head on the second toss in which case X = 2 (Corresponding to the event TH in part a))
or, we could get the first head on the third toss in which case X = 3 (Corresponding to the event TTH in part a))
or, we could get the first head on the fourth toss in which case X = 4 (Corresponding to the event TTTH in part a))
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Thus, the set of possible values of X (support of X) = {1,2,3,4,...} [Answer]
c)
d)
From part c)
We observe that the probability of the event P(X=x) has the form (1-p) multiplied by p raised to the power (x-1). Where (1-p) is the probability of observing a head and p^(x-1) denotes the probability of observing (n-1) tails.
Thus, the pmf of X is given by:
[Answer]
which we recognize as the pmf of Geometric(1-p) distribution.
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