# Probability calculations

September 10, 2004 5:41 PM Subscribe

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?

[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?

[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?

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

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

p(A or B) = p(A) + p(B) - p(A and B).

soo... p(a or b) = .15 + .05 - (.05 * .15)

on preview: ROU beat me to it.

posted by LimePi at 5:56 PM on September 10, 2004

soo... p(a or b) = .15 + .05 - (.05 * .15)

on preview: ROU beat me to it.

posted by LimePi at 5:56 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

(A && !B) + (B && !A) + (A && B) =

(0.15 * (1 - 0.05)) + (0.05 * (1 - 0.15)) + (0.15 * 0.05) =

.1425 + .0425 + .0075 = 0.1925

19.25% chance of one or more events happening.

And on preview, everybody beat me to it.

posted by mosch at 6:04 PM on September 10, 2004

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

(A && !B) + (B && !A) + (A && B) =

(0.15 * (1 - 0.05)) + (0.05 * (1 - 0.15)) + (0.15 * 0.05) =

.1425 + .0425 + .0075 = 0.1925

19.25% chance of one or more events 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

posted by lbergstr at 6:09 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

posted by LairBob at 6:17 PM on September 10, 2004

Response by poster: 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

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

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

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).

A discussion of conditional probabilities and Bayes' Theorem (extra credit).

posted by ikkyu2 at 11:09 AM on September 11, 2004

A discussion of conditional probabilities and Bayes' Theorem (extra credit).

posted by ikkyu2 at 11:09 AM on September 11, 2004

This thread is closed to new comments.

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

z = 1 / 5 for possibility w

w = 100 / 100 for possibility w

v = x + y

soo, I get:

z = (1 / 5) * (100 / 100) = 1 / 5

then:

x = (15 / 100) * (1 / 5) = 3 / 100

y = (5 / 100) * (1 / 5) = 1 / 100

and:

u = 3/100 + 1/100 = 4/100

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