(no subject)
Dec. 1st, 2003 12:38 amTwo questions:
Anyone who understands probability vis a vis dice: what is the chance you'll get a result of only one or two on three dice? What is the chance you'll get at least once six?
And has anyone ever heard of a deep fried turkey exploding in a kitchen? I presume this is a procedure that is not approached lightly.
Anyone who understands probability vis a vis dice: what is the chance you'll get a result of only one or two on three dice? What is the chance you'll get at least once six?
And has anyone ever heard of a deep fried turkey exploding in a kitchen? I presume this is a procedure that is not approached lightly.
no subject
Date: 2003-11-30 09:41 pm (UTC)no subject
Date: 2003-11-30 09:47 pm (UTC)The chance of rolling at least one '6' is the inverse of the chance of rolling no sixes. That's 5/6 x 5/6 x 5/6 (1-5 on the first die, 1-5 on the second die, 1-5 on the third die), or .5787. The inverse of that is 1-.5787, or .4213, or 42.13%.
no subject
Date: 2003-12-01 04:48 am (UTC)no subject
Date: 2003-12-01 04:56 am (UTC)Getting that on all three dice = 1/3 x 1/3 x 1/3 = 1/27 = .037 = 3.7%.
no subject
Date: 2003-12-01 04:57 am (UTC)no subject
Date: 2003-12-01 04:59 am (UTC)no subject
Date: 2003-12-01 07:09 am (UTC)However, viz. turkey, there have been several news items on fark.com of late regarding deep-fried turkey mishaps (sometimes involving devices specifically made for the purpose, which is even weirder -- must be an American thing). I had no idea such a concept existed, and between that and the turducken, I'm scarred, scarred for life. I imagine that anything that has a sizable air pocket and is then deep fried would certainly be dangerous. Turkey shrapnel! It's what's for dinner!
I've heard good things about brining turkey, mind you, which seems almost as weird.
no subject
Date: 2003-12-01 05:02 pm (UTC)no subject
Date: 2003-12-01 07:59 am (UTC)Let X be the number of times that you get a six on a fair die, upon rolling said die three times.
X ~ bin(3,1/6) => X has a Binomial Distribution with the number of trials (n) equal to 3 and the probability of success (p) equal to 1/6.
P(X=x) = (n) * p^x * (1-p)^(n-x)
(x)
Therefore, the probability that you'll get at least one six is:
1 - P(X=0)
=> 1 - [(3) * (1/6)^0 * (5/6)^3] {The funny 3 over 0 is the combination "3 choose 0"}
(0)
=> 1 - [(3!/(0! * 3!) * 1 * (0.5787)]
=> 1 - [1 * 1 * (0.5787)]
=> 1 - 0.5787
=> 0.4213
Therefore, you have approximately 42% chance of rolling at least one six on three rolls of a fair die.
N.B. Alternately, you say that the probability of getting at least one six on three rolls of a fair die is equal to 1 minus the probability of rolling no sixes at all (the probability of doing that on each roll is 5/6):
=> 1 - [5/6 * 5/6 * 5/6]
=> 1 - 0.5787
=> 0.4213
Yeah, my stats course comes in handy after all!
:)
no subject
Date: 2003-12-01 08:48 am (UTC):)
no subject
Date: 2003-12-01 04:53 pm (UTC)