# Thread: Probability at its easiest!

1. ## Probability at its easiest!

Problem: Monster drops item 1/10 of the time. I can only have 4 monster at a time per trial/save. So if I have 4 monsters that drop 1/10 of the time... my chance of getting at least 1 item is 40% a try instead of with just 1 monster a try at 10%?

2. Originally Posted by UnknownEntity
Problem: Monster drops item 1/10 of the time. I can only have 4 monster at a time per trial/save. So if I have 4 monsters that drop 1/10 of the time... my chance of getting at least 1 item is 40% a try instead of with just 1 monster a try at 10%?
No. Calculate the probability of no drops and subtract that number from 1.

Alternatively:

X is the random variable number of drops.
X ~ Binomial(n = 4, p = 1/10)
$\displaystyle \Pr (X \geq 1) = 1 - \Pr(X = 0) = 1 - \left( \frac{9}{10}\right)^4$.

3. Originally Posted by mr fantastic
No. Calculate the probability of no drops and subtract that number from 1.

Alternatively:

X is the random variable number of drops.
X ~ Binomial(n = 4, p = 1/10)
$\displaystyle \Pr (X \geq 1) = 1 - \Pr(X = 0) = 1 - \left( \frac{9}{10}\right)^4$.
Ahhh, so my Probability of getting 1 item would be somewhere around 34%, not 40%. Thanks for the help!