91影视

In a Bernoulli trial, the likelihood of the number of successive failures before a success is obtained is represented by a geometric distribution, which is a sort of discrete probability distribution. A Bernoulli trial is a test that can only have one of two outcomes: success or failure.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of California resident who does not have adequate earthquake supplies.

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand so,

$X=1,2,3,4,......$

The random variable X refers to the number of trials before the first success. Each trial is independent of others and has similar probability of success.

This implies that random variable X follows Geometric Distribution.

Thus, the distribution of X is$X~G(0.70)$

The probability that we must survey just one or two residents unless we find a California resident who does not have adequate earthquake supplies is as follow:

$$\begin{gathered} P(X=1orX=2)=P(X=1)+P(X=2) \\ {=\left[\right(1-0.70)}^{1-1}脳(0.70)]+[{(1-0.70)}^{2-1}脳(0.70)] \\ =0.70+0.21 \\ =0.91 \end{gathered}$$

The probability that we must survey at least three California residents unless we find a California resident who does not have adequate earthquake supplies is as follow:

$$\begin{gathered} P(X鈮�3)=P(X=1)+P(X=2)+P(X=3) \\ =\underset{x=1}{\overset{3}{鈭�}}{(1-0.70)}^{x-1}脳0.70 \\ =0.063 \end{gathered}$$

The number of California residents we expected to survey until we find a California resident who does not have adequate earthquake supplies is as follow:

The expected value of geometric distribution is:

$E\left(X\right)=\frac{1}{p}$where $p=0.70$

Thus,

$$\begin{gathered} E\left(X\right)=\frac{1}{p} \\ E\left(X\right)=\frac{1}{0.70} \\ E\left(X\right)=1.428 \end{gathered}$$

The number of California residents we expected to survey until we find a California resident who does have adequate earthquake supplies is as follow:

The expected value of geometric distribution is:

$E\left(X\right)=\frac{1}{p}$where, $p=0.30$

Thus,

$$\begin{gathered} E\left(X\right)=\frac{1}{0.30} \\ E\left(X\right)=3.33 \end{gathered}$$