Respuesta :
Answer:
The expected number of people you will have to ask before you find someone with jumper cables is 6.25.
Step-by-step 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:
    [tex]p_{X}(k)=(1-p)^{k-1}p;\ k={1, 2, 3...}[/tex]
- 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:
    [tex]p_{X}(k)=(1-p)^{k}p;\ k={0, 1, 2, 3...}[/tex]
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:
[tex]E(X)=\frac{1}{p}\\\\=\frac{1}{0.16}\\\\=6.25[/tex]
Thus, the expected number of people you will have to ask before you find someone with jumper cables is 6.25.