Consider a sequence of independent Bernoulli trials with p= 0.2
a) what is the expected number of trials to obtain the first success?
for this I thought of it as geometric so 5 trials
b) after the eighth success occurs, what is the expected number of trials to obtain the ninth success?
I don't know how to go about this one?
a) 1/.2
To find the expected number of trials to obtain the first success in a sequence of independent Bernoulli trials with a success probability of p = 0.2, you can use the property of the geometric distribution.
a) Expected number of trials to obtain the first success:
The geometric distribution represents the number of trials required to achieve the first success in a sequence of independent trials. The expected value of a geometric distribution is given by 1/p, where p is the probability of success. In this case, p = 0.2, so the expected number of trials to obtain the first success is 1/0.2 = 5 trials.
b) Expected number of trials to obtain the ninth success after the eighth success occurred:
To find the expected number of trials to obtain the ninth success, given that the eighth success has already occurred, you can consider the remaining sequence of trials after the eighth success and apply the same logic.
In this case, you need to find the expected number of trials to obtain the first success in a sequence of independent Bernoulli trials with p = 0.2. The expected value of the geometric distribution is still 1/p, but it represents the number of trials required to achieve the first success after the eighth success.
Since the eighth success has already occurred, you have a sequence of independent trials starting from the next trial. So, the expected number of trials to obtain the ninth success is also 1/0.2 = 5 trials.
Therefore, after the eighth success occurs, the expected number of trials to obtain the ninth success in a sequence of independent Bernoulli trials with p = 0.2 is also 5 trials.