Question

Forahigh-riskdriver,thetimeindaysbetweenthe beginning of a year and an accident has an exponential pdf. Suppose an insurance company believes the probability that such a driver will be involved in an accident in the first forty days is 0.25. What is the probability that such a driver will be involved in an accident during the first seventy-five days of the year?

Answer 1

Answer:

The probability that such a driver will be involved in an accident during the first seventy-five days of the year is 0.408

Step-by-step explanation:

Consider the provide information.

The probability that such a driver will be involved in an accident in the first forty days is 0.25.

Therefore, $P(X\leq 40)=0.25$

Where is X represents the random variable that shows time between beginning of a year and an accident.

X is the exponential distribution with λ as a parameter.

$P(X\leq x)=1-e^{-\lambda x}, x\geq 0$

Therefore,

$P(X\leq 40)=1-e^{-40\lambda}$

$1-e^{-40\lambda}=0.25$

$e^{-40\lambda}=0.75\\-40\lambda=ln(0.75)\\\lambda=0.00719\approx0.007192$

Now we want the probability that a driver will be involved in an accident during the first seventy five days.

$P(X\leq 75)=1-e^{-75\times 0.0072}=0.408$

Hence, the probability that such a driver will be involved in an accident during the first seventy-five days of the year is 0.408