Calculating Chance: odds of drawing desired number on 1st draw of 2 of 6 balls in bag

[Smitty517]

For whatever reason, I cannot wrap my head around this at 5:30 in the morning. I registered just to post about it and get it out of my head.

I was proposed a situation where there are 6 items available, we could say they are numbered balls in a bag. It is said that two are chosen at random, and it's impossible to pick the same one twice. Your goal is to get a specific individual number within two picks. You can preform this an unlimited number of times to get your number, but I want to know how the odds are of getting it on the first try.

The simple answer feels like it's a 1 in 3 chance, but I have a feeling that's not nearly right considering how probability works, but if someone could explain why this is or is not the right answer, I would appreciate it.


Thanks,
Smitty
~{no sig yet}~

 

[Smitty517]

Calculating Chance

For whatever reason, I could not wrap my head around this at 5:30 in the morning. I registered just to post about it and get it out of my head.

I was proposed a situation where there are 6 items available, we could say they are numbered balls in a bag. It is said that two are chosen at random, and it's impossible to pick the same one twice. Your goal is to get a specific individual number within two picks. You can preform this an unlimited number of times to get your number, but I want to know how the odds are of getting it on the first try.

The simple answer feels like it's a 1 in 3 chance, but I have a feeling that's not nearly right considering how probability works, but if someone could explain why this is or is not the right answer, I would appreciate it.


Thanks,
Smitty
~{no sig yet}~

 

[ksdhart2]

Okay, so if I'm understanding your scenario correctly, you have 6 balls in a bag, numbered 1-6, and you pick one of these balls. Then, without putting that ball back, you pick another ball. And you want to know what the odds of one of these balls having some random number (call that n maybe). If that's correct, here's a few questions you might ask yourself to determine whether your intuition of 1/3 was the correct chance or not. For simplicity's sake, let's say n = 2. On the first ball, what is the chance that you'll draw a 2? Assuming you didn't get a 2, what's the chance the second ball will be a 2? Now that you know the chance of the first ball being 2 and the chance of the second ball being 2, what's the chance of either one being the 2? Finally, does the value of n matter? Why or why not?

 

[pka]

(T)here are 6 items available, we could say they are numbered balls in a bag. It is said that two are chosen at random, and it's impossible to pick the same one twice. Your goal is to get a specific individual number within two picks. You can preform this an unlimited number of times to get your number, but I want to know how the odds are of getting it on the first try.

Actually the question as stated makes no real sense.
The probability of getting the ball on the first draw is one out of six.
If two balls are drawn, without replacement, the probably that the ball is one of those two is one out of three.
Given that the ball is one of the first two drawn, the probably that the ball was the first drawn is one out of two.

But from the way you posed the question, I have no idea which you meant.