If the likelihood of event A occurring is 15%, and the likelihood of independent event B occuring is 5%, what is the likelihood of EITHER event occuring? posted by sohcahtoa to (13 comments total)
[StatisticsFilter...]So, I know the answer isn't 20%, and I'm pretty sure it's not 15%. I'm sure there's a formula that could be applied, but I was an english major.
I reasoned as far as thinking of two coin tosses - the likelihood of EITHER coin toss yeilding HEADS is greater than 50% but less than 100%. But that's where I stopped.
Incidentally, this came up from reading the normally stats-savvy Sons of Sam Horn discussion board. In it someone opines that there's a 5% chance of Ichiro hitting .400 this year and a 15% chance of Bonds hitting .400 this year. The author then makes an aside of there being a "1 in 5" chance of there being a .400 hitter this year, which I think is incorrect.
Unfortunatly, much like Metafilter, as far as I can tell SOSH is a closed forum at this point, so I can't ask directly. Anyone wanna get me in? posted by sohcahtoa at 5:42 PM on September 10, 2004
How do you know it's not 20%? Seems that mathematically, all possibilities should add up to 100%, so it makes sense that you would just add the possibilities. It works the same in lotteries. Having a 1 in a million chance and getting 2 tickets doesn't mean you get twice the chance of winning (1/2), it means you have 2 / 1,000,000 chances of winning.
In the case you have, though, your problem is compounded. Your looking at chances OF a chance, not chances WITHIN a chance.
x = 15 / 100 for possibility z
y = 5 / 100 for possibility z
Or, I get 4% that of the 0.400 batting averages either event will occurr.
Anyone want to tell me where I went wrong? I barely passed stats. posted by shepd at 5:53 PM on September 10, 2004
For two independent events A and B, the probablity of (A or B) is pr(A) + pr(B) - pr(A and B). It's the last part that keeps you from nonsense like there being a 150% chance of getting at least one heads in three coin-tosses -- you get that by counting the same probability mass two (or more) times.
So it's 0.15 + 0.05 - (0.15*0.05). Which seems to be 0.1925 if one can trust google calculator. posted by ROU_Xenophobe at 5:54 PM on September 10, 2004
on preview: ROU beat me to it. posted by LimePi at 5:56 PM on September 10, 2004
ROU and LimePi - Thanks! posted by sohcahtoa at 6:01 PM on September 10, 2004
0.15 = chance of event A happening
0.05 = chance of event B happening
0.15 * ( 1 - 0.05 ) = chance of only event A happening
0.05 * (1 - 0.15) = chance of only event B happening
0.15 * 0.05 = chance of both event A and event B happening
And on preview, everybody beat me to it. posted by mosch at 6:04 PM on September 10, 2004
The lottery ticket analogy only works if they're two tickets to different lotteries (otherwise they're not independent events.) posted by lbergstr at 6:09 PM on September 10, 2004
YAY. I lose. posted by shepd at 6:10 PM on September 10, 2004
I just want to point out that "sohcahtoa" is a pretty math-oriented nickname to pick, for an english major. posted by LairBob at 6:17 PM on September 10, 2004
Yes. I didn't miss that irony (or that of the misspellings in my above, for that matter).
It was a remnant from my rock-and-roll days. posted by sohcahtoa at 6:24 PM on September 10, 2004
huh. My instinct is to calculate this by figuring "p(A or B) = 1 - p(not A) * p(not B)".
So, 1-(.85 * .95) = 0.1925.
Yet another way of getting the same answer. I figure it's worth mentioning since it might sometimes be easier to estimate. Also note that the smaller the probabilities of each event, the closer p(A)+p(B) is to the right answer. So 1 in 5 isn't far off in this case. posted by sfenders at 7:36 PM on September 10, 2004
FWIW, you ought to check out Sox Therapy. SoSH suxks. posted by Kwantsar at 8:09 PM on September 10, 2004
To be extremely pedantic: the original questioner used the word 'likelihood' where 'probability' would be the correct term. 'Likelihood' is a term used to refer to conditional probabilities (non-independent events).
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I reasoned as far as thinking of two coin tosses - the likelihood of EITHER coin toss yeilding HEADS is greater than 50% but less than 100%. But that's where I stopped.
Incidentally, this came up from reading the normally stats-savvy Sons of Sam Horn discussion board. In it someone opines that there's a 5% chance of Ichiro hitting .400 this year and a 15% chance of Bonds hitting .400 this year. The author then makes an aside of there being a "1 in 5" chance of there being a .400 hitter this year, which I think is incorrect.
Unfortunatly, much like Metafilter, as far as I can tell SOSH is a closed forum at this point, so I can't ask directly. Anyone wanna get me in?
posted by sohcahtoa at 5:42 PM on September 10, 2004