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| revprev | Admin | Jun 26'24 at 16:15 | −14 | m | |
| revcurprev | Admin | Jun 26'24 at 16:10 | −16 | m | |
| revcurprev | Admin | Jun 26'24 at 16:07 | +64 | m | |
| revcur | Admin | (Created page with "Let <math>T</math> equal the number of tosses until the first head appears. <math>T</math> has a geometric distribution with <math>P(T=k) = (1/2)^k </math> for <math>k\geq 1 </math>. The probability that the first head appears on the fifth toss, given that it has not appeared in the first two tosses, equals <math display = "block"> P(T=5|T>2) = \frac{P(T=5)}{P(T>2)} = \frac{(1/2)^5}{\sum_{k \geq 3} (1/2)^k } </math>") | Jun 26'24 at 16:06 | +420 |