Suppose a biased coin comes up head with a probability of 0.2 when tossed. What is the probability that you do not flip heads until the 1st, 2nd, 3rd,and 4th tosses?
Let the probability of success, p = 0.2.
We are looking for the probability that you do not flip heads until the first toss, which means that we are looking for the probability that there are exactly 0 failures before the first success:
P(0 failure before 1st success) = f(0) = P(X = 0) =
$\left(0.2\right){\left(1-0.2\right)}^{0}$ = ${0.20}$
P(1 failure before 1st success) = f(1) = P(X = 1) =
$\left(0.2\right){\left(1-0.2\right)}^{1}$ = ${0.16}$${}$
P(2 failures before 1st success) = f(2) = P(X = 2) = $\left(0.2\right){\left(1-0.2\right)}^{2}$ = ${0.128}$
P(3 failures before 1st success) = f(3) = P(X = 3) = $\left(0.2\right){\left(1-0.2\right)}^{3}$ = ${0.1024}$