Answer:
The expected number of people you will have to ask before you find someone with jumper cables is 6.25.
Explanation:
- The geometric distribution is the distribution of the number of X Bernoulli trials required for one success. If the one success requires k independent trials, each with the probability of success as p, then the probability that the kth trial is the one success is:
- The geometric distribution is the distribution of the number of Y = X – 1 failures required before the first success. If the first success requires k failures, each trial with the probability of success as p, then the probability that the 1st success occurs after k failures is:
![p_(X)(k)=(1-p)^(k)p;\ k={0, 1, 2, 3...}](https://img.qammunity.org/2021/formulas/mathematics/college/javp28vaxohdoefgb6lcxvddcm8zqd8b0v.png)
In this case the first scenario follows.
X = number of people you will have to ask before you find someone with jumper cables.
p = 0.16
Compute the expected number of people you will have to ask before you find someone with jumper cables as follows:
![E(X)=(1)/(p)\\\\=(1)/(0.16)\\\\=6.25](https://img.qammunity.org/2021/formulas/mathematics/college/tq9taq738k247riyvu61n9u5uyo1finvjh.png)
Thus, the expected number of people you will have to ask before you find someone with jumper cables is 6.25.