Binomial Distribution on finding the potency of a drug

Question

An experiment is designed to test the potency of a drug on 40 rats. Previous animal studies have shown that a $10$-mg dose is lethal $10$% of the times within the first $4$ hours.

What is the probability that between $2$ and $8$ rats die during the experiment due to the drug?

My attempt: Let $X$ be the number of rats that die in the first 4 hours. $$P(2 \leq x \leq8) = \sum_{x=2}^{8} \binom{40}{x} (0.1)^{x} (0.9)^{40-x}$$

To make the drug more potent, the company came with a new formula. This reduced the chances of a new drug being lethal to $1$%. The new drug is administered to $10000$ rats. Approximate the exact probability that $5$ rats die.

My attempt:

$$P(x=5) = \binom{10000}{5} (0.01)^{5} (0.99)^{10000-5}$$

Am I on the right path?

Answer 1

In order to calculate the first statement

$$ P(2 \leq X \leq 8) = \sum_{x=2}^{8} \binom{40}{x} \big(0.1\big)^{x} \big(0.9\big)^{40-x} \tag{1}$$

We can utilize some small python script as it is kind of hard by hand.

from scipy import stats

# number of rats
n=40
# lethal dose probability
p=.1
# upper bounds
kmax = 8
kmin = 1
# cdf summations
y1 = stats.binom.cdf(kmin, n, p)
y2 = stats.binom.cdf(kmax,n,p)
# differences
y = y2-y1
y
   y
Out[13]: 0.9040309985710783

Note the binom.cdf command is

$$ F(x|n,p) = \sum_{i=0}^{n} \binom{n}{i} \big(p)^{i} \big(1-p\big)^{n-i} \tag{2}$$

So we to set our limits

$$ I_{1} = F(x|1,0.1) = \sum_{i=0}^{1} \binom{1}{i} \big(0.1)^{1} \big(0.9 \big)^{1-i} \tag{3}$$

$$ I_{2} F(x|8,0.1) = \sum_{i=0}^{8} \binom{8}{i} \big(0.1)^{8} \big(0.9 \big)^{8-i} \tag{4}$$ then we find

$$ I = I_{2} - I_{1} \tag{5} $$ there is a 90% chance $2$ - $8$ rats die.

ok..the second one

$$ P(X=5) = \binom{10000}{5} \big(0.01\big)^{5} \big(0.99\big)^{9995} \tag{6}$$

from scipy import stats

# number of rats
n=10000
# lethal dose probability
p=0.01
# exact number 
k = 5
# pmf 
y = stats.binom.pmf(k, n, p)
y
Out[8]: 1.968586020940221e-36

Wow, startling difference.. I don't think you can do these by hand. You can make rough approximations.

In the last equation you get some rough terms...

$$ P(X=5) = \frac{10000!}{5!9995!} \frac{1}{100^{5}} \frac{99^{9995}}{100^{9995}} \tag{7}$$ Various things will drop out

$$ P(X=5) = \frac{10000\cdot 9999 \cdot 9998\cdot 9997\cdot 9996 }{120} \frac{1}{100^{2} \cdot 10^{3}} \frac{99^{9995}}{100^{9995}} \tag{8}$$

$$ P(X=5) = \frac{ 9999 \cdot 9998\cdot 9997\cdot 9996 }{120} \frac{1}{ 10^{3}} \frac{99^{9995}}{100^{9995}} \tag{9}$$

It just becomes a really small number..this is the exact reason calculators were invented.