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Ahhh. Dumb probability question

five40

Golden Member
This is a really dumb question. I haven't been in school for a good number of years and I can't remember the formula for a simple probability problem. For example if I have four coins, what are the odds that at least 2 of them are tails after 1 flip of each coin. Now I know the answer is 11/16 but I can't figure out the middle part. By simple counting I know that:

Chances of getting exactly 0 tails out of 4 is 1/16
Chances of getting exactly 1 tail out of 4 is 4/16
Chances of getting exactly 2 tails out of 4 is 6/16
Chances of getting exactly 3 tails out of 4 is 4/16
Chances of getting exactly 3 tails out of 4 is 1/16

The thing I can't remember is how to get the 4/16, 6/16 mathematically. Yeah it's real easy to count for small number sets, but what if I'm flipping 400 coins?
 
Speaking of probability I was sitting on the crapper at work and started thinking, what are the chances of someone coming in to take a dump next to me?
 
Originally posted by: Soccer55
nCr * p^r * (1-p)^(n-r)

Where:
n = number of trials
r = number of successes
p = probability of success

-Tom

Does that work for my example? I'm probably doing something wrong:

(n*r)=(4*2)
p^r=(1/2)^4
(1-p)^(n-r)=(1-1/2)^(4-2)

nCr * p^r * (1-p)^(n-r) = 8 * .0625 * .25 = .125

However by counting we know that getting 2 tails out of 4 is 6/16 which is .375
 
You want 2 successes out of 4 trials so n=4, r=2.

nCr is not simply n*r. nCr = n!/[r! * (n-r)!]. So 4C2 would be 4!/[2! * (4-2)!] = 4!/(2! * 2!) = 24/4 = 6.

So you have 6 * (1/2)^2 * (1/2)^(4-2) = 6 * (1/2)^2 * (1/2)^2 = 6 * (1/2)^4 = 6/16 = 3/8 = .375

-Tom
 
Originally posted by: Soccer55
You want 2 successes out of 4 trials so n=4, r=2.

nCr is not simply n*r. nCr = n!/[r! * (n-r)!]. So 4C2 would be 4!/[2! * (4-2)!] = 4!/(2! * 2!) = 24/4 = 6.

So you have 6 * (1/2)^2 * (1/2)^(4-2) = 6 * (1/2)^2 * (1/2)^2 = 6 * (1/2)^4 = 6/16 = 3/8 = .375

-Tom

Thanks!
 
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