Answer:
Step-by-step explanation:
From above-given information
Consider p to be the probability that the first coin comes up as head
Consider q to be the probability that the second coin comes up as tail.
Provided all tosses seems to be independent.
Then;
Let X represent no. of tosses, where X follows a geometric distribution with parameter;
p' = p(1-q) +q(1-p)
with probability mass function (pmf)
[tex]P(X = x) = ( 1 - p')^{x - 1} p'[/tex]
Thus, the mean (expected value) and the standard deviation can be computed as:
[tex]E(X) = \dfrac{1}{p'}[/tex]
[tex]\mathbf{E(X) = \dfrac{1}{p(1-q)+q(1-p)}}[/tex]
[tex]V(X) = \dfrac{1-p'}{(p')^2}[/tex]
[tex]\mathbf{V(X) = \dfrac{1-[p(1-q)+q(1-p)]}{[p(1-q)+q(1-p)]^2}}[/tex]
(b)
Suppose X represents that the first coin is head;
and Y represent the last toss
Therefore, the required probability is:
[tex]P(X|Y) = \dfrac{p(1-q)}{p'}[/tex]
[tex]P(X|Y) = \dfrac{p(1-q)}{p(1-q)+(1-p)q}[/tex]
