What is the probability for the second half of the question posted?

BGM

TS Contributor
#2
(no offense) Actually I think the blog just complicated the matter.

If you insist to follow his way, just note that there are 6 + 6 + 6 + 2 = 20 possible cases. The order following his configurations labels. Each cases are equally likely.

For the first part:
You randomly select a slot, and there will be 3 + 3 + 3 + 1 = 10 possible cases which the selected slot is blank, such that it satisfy the first condition. Again each cases are equally likely. Count carefully, there will be 1 + 2 + 2 + 1 = 6 cases that the next slot contain a bullet (assuming it is rotating in some direction). So the required probability is 6/10 = 3/5. Remember conditional probability is just another probability, which focus on a shrink sample space. Maybe the blog is counting wrong or I miss some point here.

For the second part:
There will be 2 + 1 + 1 + 0 = 4 possible cases satisfying the condition (consecutive blank) and only 1 + 0 + 0 + 0 = 1 case out of them satisfying the next slot is also blank. So the probability is 1/4.

A much more straight forward way:
Consider an urn with 3 black balls and 3 non-black balls. Balls are drawn without replacement.

a) What is the probability that the 2nd drawn is black given the first drawn is non-black?
b) What is the probability that the 3rd drawn is non-black given the first two drawns is non-black?