Question

A dairy farmer accidentally allowed some of his cows to graze in a pasture containing weeds that would contaminate the flavor of the milk from this herd. The farmer estimates that​ there's a a 9 9​% chance of a cow grazing on some of the flavorful weeds. ​(a) Under these​ conditions, what is the probability that none of the 16 16 animals in this herd ate the tasty​ weeds? ​(b) Does the Poisson model give a good estimate of the probability that no animal ate the​ weed?

Answer 1

Answer:

the required probability of is 0.1886

the approximate probability is 0.2052

Step-by-step explanation:

The farmer estimates that​ there's a a 9 9​% chance of a cow grazing on some of the flavorful weeds

i.e P = 9.9% = 0.099

Let assume that X is a description of how the cows are grazing on some of the flavorful weeds.

The probability density function of the binomial distribution is :

$\mathbf{P(X=x)=(^n_x)_{p^x}(1-p)^{n-x}}$

a)

To calculate that the probability that none of the 16  animals in this herd ate the tasty​ weeds.

$\mathbf{P(X=0)=(^{16} _0){(0.099)^0}(1-0.099)^{16-0}}$

= $\mathbf{1*1*0.1886}$

= 0.1886

Thus; the required probability of is 0.1886

b) To calculate the probability that no animal ate the weed.

By using Poisson approximation model:

$\mathbf{P(X=0) = \frac{e^{-(np)}(np)^x}{x!} }$

$\mathbf{P(X=0) = \frac{e^{-(16*0.099)}(16*0.099)^0}{0!} }$

$\mathbf{P(X=0) =e^{-1.584}}$

$\mathbf{P(X=0) =1.5254*10^{-7}}$

               = $\mathbf{0.2052}$

Hence; the approximate probability is 0.2052