Yeah for my side!

[BigV]

Originally Posted by HungLikeJesus

I was at a work meeting and five us ended up at lunch together. Of the five, three were left handed. Immediately after lunch I calculated the odds of that happening. Anybody want to guess at the answer? (The calculation involves the binomial theorem.)

3/5?

[HungLikeJesus]

Originally Posted by BigV

3/5?

I might be wrong, but I get 1.05%

Probability of being left-handed = 11%,
Probability of exactly 3 out of 5 being left-handed (randomly chosen) =
(0.11^3)*(0.89^2)*(5!)/(3!*2!) = 0.0105

Of course, there are problems with the logic (e.g. how do you define left-handed?).