91Ó°ÊÓ

Probability is just the way in which likely something is to occur. At the point when we're uncertain about the result of an occasion, we can discuss the probabilities of specific results — how likely they are. The investigation of occasions represented by likelihood is called insights.

We know,

Total number of years the earthquake occurred =$14$

$=14×12=168$months

The number of earthquakes that occurred in$168$months = $29$

One earthquake will occur in$\mathrm{μ}$= $\frac{168}{29}=5.79\text{months}$

Now, $m=\frac{1}{\mathrm{μ}}=\frac{1}{5.79}=0.1727$

x = $3$months

Using,

$P(X<x)=1−e^{−mx}$we find the probabilty that the next earthquake occurs within the next three months

$\begin{array}{r}P(x<3)=1−e^{−0.1727×3}\end{array}$

$\mathrm{P}(\mathrm{x}<3)=1−0.5957=0.4043$

We know,

m = $0.1727$

the probability that the next three months will be free of earthquakes after six months passed is,

$\mathrm{P}(6<x<9)$

Using,

$$\begin{gathered} \mathrm{P}(6<x<9)=1−\mathrm{P}(x<9)−P(x<6) \\ =1−\left(e^{−0.1727×9}−\left(1−e^{−0.1727×6}\right)\right) \\ =1−(0.3548−0.2133)=0.8585 \end{gathered}$$

We know,

m = $0.1727$

The number of months is x =$12$

The probability of zero earthquakes occurring in a given year or $12$months is

$P(X<x)=1−e^{−mx}$

$$\begin{gathered} 1−\mathrm{P}(x<12)=1−\left(1−e^{−0.1727×12}\right) \\ 1−\mathrm{P}(x<12)=e^{−0.1727×12} \\ =0.1259 \end{gathered}$$

we know,

The probability of $29$earthquakes in $168$months is

role="math" localid="1652277871482" $\mathrm{μ}=\frac{29}{168}=0.1726$

now, the decay rate

$$\begin{gathered} m=\frac{1}{\mathrm{μ}} \\ m=\frac{1}{0.1726}=5.79 \end{gathered}$$

the probability that at least two earthquakes will occur is,

$$\begin{gathered} \mathrm{P}(x>2)=1−\mathrm{P}(x<2) \\ \mathrm{P}(x>2)=1−\left(1−e^{−5.79×2}\right) \\ =0.000009 \end{gathered}$$