# Thread: Another probability question

1. ## Another probability question

I am having difficulty trying to get correct answer for the following problem:

Mr. Smith has 8 master keys that unlock several chests. Only one master key will unlock a given chest. 40% of these chests are usually left open, what is the probability that Mr. Smith can unlock a specific chest if he selects 3 master keys at random?

My attempt: I have a feeling this maybe totally wrong.

I am pretty sure it is a binomial distribution problem

P(X=3) = (8C3)(.4^3)(.6^5) = 3.66176

2. There's a problem just like that in walpole/meyers.
You need to break it into two cases, the chest being open or locked.
So it's
(.4) + (.6)(3/8)
The .4 is your chance that the chest is open.
The second pair of numbers is the chance it's locked times the chance
you selected a key that will open it.
In walpole they use doors to homes....

3. Originally Posted by mathwiz2006
I am having difficulty trying to get correct answer for the following problem:

Mr. Smith has 8 master keys that unlock several chests. Only one master key will unlock a given chest. 40% of these chests are usually left open, what is the probability that Mr. Smith can unlock a specific chest if he selects 3 master keys at random?

My attempt: I have a feeling this maybe totally wrong.

I am pretty sure it is a binomial distribution problem

P(X=3) = (8C3)(.4^3)(.6^5) = 3.66176

Yikes, a probability larger than 1?