Respuesta :
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, [tex]P(X\leq 40)=0.25[/tex]
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.
[tex]P(X\leq x)=1-e^{-\lambda x}, x\geq 0[/tex]
Therefore,
[tex]P(X\leq 40)=1-e^{-40\lambda}[/tex]
[tex]1-e^{-40\lambda}=0.25[/tex]
[tex]e^{-40\lambda}=0.75\\-40\lambda=ln(0.75)\\\lambda=0.00719\approx0.007192[/tex]
Now we want the probability that a driver will be involved in an accident during the first seventy five days.
[tex]P(X\leq 75)=1-e^{-75\times 0.0072}=0.408[/tex]
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
