Probability of Finding Shrubs in Jojoba Acres

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4)An agronomist has found an average of 2 shrubs per acre of jojoba, which serves as the source for transmission oil. Find the Following probabilities:
a) That 5 plants will be found when the number of acre search is (1) one; (2) two;
or (3) three.
b) That no plants will be found in searching the next (1) .5 acre; (2) 1 acre; or (3)
1.5 acres

I want to just do this, but I don't think I can:

The Agronomist finds an average of 2 shrubs per 1 acre jojoba. μ = 2

The probability is given by: μ = np, hence: p = (μ/n)

a.) (1) p = (2/5) = 0.4 (2) p = (2)/(5/2) = 0.8 (3) p = 1.2


120% does not seem rational. If its going to happen, it should be 100% at most.
 
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Hi!

First things first,definitions! Let X be the random variable for the number of shrubs per acre of jojoba.

In my opinion, X should follow a Poisson distribution, as there is theoretically no limit to the number of shrubs we can find, instead of a Binomial model as seen in your calculations. Hence, X \sim P_0(2)
With this, you should be able to proceed with your working for (a).

As for part (b), I think we should pay some attention to the word "next". This probably means that there is some restriction on the probability. I think the question means that the agronomist should find 5 plants in the first area of land... but which area (1, 2, or 3 acres) should we use?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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