While traveling to Europe, Phelan exchanges 250 US dollars for euros. He spent 150 euros on his trip. After returning to the United States he converts his money back to US dollars. How much of the original 250 US dollars does Phelan now have?

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grayfaith16

08.10.2022

Mathematics

Probability of more than 1 death in a corps in a year

12.52%

Probability of no deaths in a corps over 7 years

1.40%

Step-by-step explanation:

If the data collected follow a Poisson distribution with mean 0.61, then

Probability of k deaths in a year is

$P(k\;deaths\;in\;one\;year)=\frac{0.61^ke^{-0.61}}{k!}$

The probability of more than 1 death in a corps in a year would be

(1) $P(2)+P(3)+...=\sum_{k=2}^{\infty}P(k)=\sum_{k=2}^{\infty}\frac{0.61^ke^{-0.61}}{k!}=e^{-0.61}\sum_{k=2}^{\infty}\frac{0.61^k}{k!}$

But the Taylor series around 0 for the exponential function is

$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$

So,

$e^{0.61}=\frac{0.61^0}{0!}+\frac{0.61^1}{1!}+\sum_{k=2}^{\infty}\frac{0.61^k}{k!}$

and

$\sum_{k=2}^{\infty}\frac{0.61^k}{k!}=e^{0.61}-1.61$

Replacing in (1), we obtain

$\sum_{k=2}^{\infty}P(k)=e^{-0.61}(e^{0.61}-1.61)=1-1.61e^{-0.61}=1-0.874794\approx 0.1252$

or 12.52%

The probability of 0 deaths in one year is

$P(0)=e^{-0.61}$

As the events are independent, the probability of 0 deaths over 7 years is the product of P(0) by itself 7 times:

$P(0))^7=(e^{-0.61})^7=e^{-4.27}\approx 0.0140$

or 1.40%

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