# Need help with probability homework! (Bayesian model)?

1. Suppose a pill exists with the property that one out of every 100 such pills contains a lethal amount of a substance, and that the other 99 are harmless.

a. Calculate how many people would be expected to die if:

a 1000 people each take 1 pill,

if 1000 people each take 2 pills,

and if 1000 people each take 100 pills.

Assume each person is taking the pills from a huge source of pills, so that the “1 out of 100” statistic remains unchanged as the pills are taken.

Okay, for the first one, I assumed 1/100 chance of death. Since its taken with replacement, the answer = 1/100 * 1000 people = 10 people will die.

For the second one, Probability of getting pill on first try OR second try. P(A) or P(B) = P(A) + P(B) = 1/100 + 1/100 = 2/50 * 1000 = 20 people will die.

What happens if everyone takes 100 pills?! It seems like everyone is going to die!

Update:

Thank you so much! I actually did get 19.9 deaths for 2 pills, but I thought it was wrong, so I started to do it another way.

(:! Thanks again!

-Johnny

Relevance
• Mitch
Lv 7

Hi Johnny,

Kudos to you for the proper use of Y!A in seeking help with your homework.

I can see you've attempted the problem and followed up with good logic.

Because of this, I'm going to afford you the benefit of solving this problem.

(It's normally something I DON'T do).

When you get into multiple independent probabilities, the formula changes!

"If 1000 people take two pills... how many will die?"

Let's look at the problem from a different angle... "How many will NOT die?"

P(alive after taking two pills) = (.99) * (.99) * (1000 people) = 980.1 Living Souls.

This means on average, 19.9 die from taking this medication.

(Notice it's not exactly the 20 deaths you expected)!

P(alive after taking 100 pills) = (.99)^100 * (1000 people) = 366.032 Survivors.

This means that about 634 people are expected to perish from taking this drug.

Does that make sense?