Question for quant/stat geeks

[DCL]

let's say there are 3 baseball players. Player A bats .200, Player B bats .300, and Player C bats .275

what is the probability that at least one of these players will get a hit on their first at-bat?

actually, i don't need the answer. just want to know the formula to calculate something like this.

[ace23]

Binomial distribution.

Hold up did it wrong

[fiddy]

nvm

[DeuceWallaces]

I'm a modeling stats guy, didn't like probability too much, but I believe it's.

P(A) = 1 - (0.8 x 0.7 x 0.725) = 0.594

[Jailblazers7]

I'm a modeling stats guy, didn't like probability too much, but I believe it's.

P(A) = 1 - (0.8 x 0.7 x 0.725) = 0.594

Yep this is right. It just 1 minus the probability of all 3 batters not getting a hit.

[KevinNYC]

I'm a modeling stats guy, didn't like probability too much, but I believe it's.

P(A) = 1 - (0.8 x 0.7 x 0.725) = 0.594

Why are you using the decimal for failure (didn't get a hit) and not for success (did get a hit)?

what is the probability that at least one of these players will get a hit on their first at-bat?

[DeuceWallaces]

Because at one time I had to take the GRE. And this is the fast way.

Notice the 1-

[ace23]

Yep this is right. It just 1 minus the probability of all 3 batters not getting a hit.
It is not right. He said at least one hit, so you must take into the accounts of 1, 2, and 3 hits.

Answer is Binomial(1,3,.25833) + Binomial(2,3,.25833) + Binomial(3,3,.25833) = 0.59202632798

[DeuceWallaces]

It is right. It's the fastest way to do it. It's 1 minus the probability of not getting a hit.

[Jailblazers7]

It is not right. He said at least one hit, so you must take into the accounts of 1, 2, and 3 hits.

Answer is Binomial(1,3,.25833) + Binomial(2,3,.25833) + Binomial(3,3,.25833) = 0.59202632798

I'm a little confused. Doesn't the probability of at least 1 hit contain the probability that there could be multiple hits? It's been a long time since I've done probabilities so I could be forgetting some things.

[ace23]

I'm a little confused. Doesn't the probability of at least 1 hit contain the probability that there could be multiple hits? It's been a long time since I've done probabilities so I could be forgetting some things.
Well I guess the way he did it makes sense, and we got close to the same thing. I'm sure my answer is right though. Let me think a bit.

[DeuceWallaces]

No need to think on it. They're the same thing. My was is faster, which is why they teach it for the GRE and GMAT. Technically mine is more correct, because it appears you rounded the 31/120 whereas I took the original averages.

[Im Still Ballin]

PM me if you want to know a little more

[DeuceWallaces]

Lol and the bitch negged me. Who's getting shit on now? Internet Rapper.

[ace23]

No need to think on it. They're the same thing. My was is faster, which is why they teach it for the GRE and GMAT. Technically mine is more correct, because it appears you rounded the 31/120 whereas I took the original averages.
Yeah, it's the same thing. Forgot that you can do that with "at least 1" scenarios. I over-complicated it.

And I didn't neg you. Lol.

[DCL]

i repped both of you

[KevinNYC]

Because at one time I had to take the GRE. And this is the fast way.

Notice the 1-

Ah. I did not notice the one minus at first. I think I was surprised it was as high as 60%, but if it makes sense. If they were batting .333 the probability would be 1.

Thanks.

[DeuceWallaces]

No it would be 0.704

[DCL]

corrected some messy cells and remodeled excel...

after 10,000 events, the probability was .59404, as predicted by the formula

laws of probability :bowdown:

http://111.imagebam.com/download/DTyAeIj_gLC412jJrK6hvg/39534/395331064/excel.jpg

[KevinNYC]

going to get a coffee after using addition. not multiplication.

It's been a long time since I used any of this stuff. I used to teach GED tests.

Also I should probably get more than 5 hours a sleep.

:facepalm:

[DeuceWallaces]

Kewl