# Statistics 101 **URL:** \ **Category:** Factual Questions\ **Created:** [September 8, 2001, 6:55pm UTC](https://boards.straightdope.com/t/statistics-101/80421 "2001-09-08T18:55:49Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![SuaSponte](https://avatars.discourse-cdn.com/v4/letter/s/67e7ee/32.png) [@SuaSponte](https://boards.straightdope.com/u/SuaSponte)\ **Post date:** [September 8, 2001, 6:55pm UTC](https://boards.straightdope.com/t/statistics-101/80421/1 "2001-09-08T18:55:49Z")
The sad part is I took stats around 12 years ago, and even got an ‘A’. But I just can’t remember the formula for determining odds for multiple attempts. Say you flip a coin three times. What are the odds that you will get at least one “heads” and how determine the odds? Thanks. Sua ---
**Author:** ![Dr.Lao](https://avatars.discourse-cdn.com/v4/letter/d/2bfe46/32.png) [@Dr.Lao](https://boards.straightdope.com/u/Dr.Lao)\ **Post date:** [September 8, 2001, 7:05pm UTC](https://boards.straightdope.com/t/statistics-101/80421/2 "2001-09-08T19:05:34Z")
I can’t think of the general way to do this, but it is easier to calculate the odds that you will NOT get at least one heads. You will have to get three tails so that is: P(t and t and t) = P(t)\*P(t)\*P(t) = 0.5 \* 0.5 \* 0.5 = 0.125. At least one heads must be 1 - (NOT at least one heads) so: P(at least one heads) = 1 - P(t and t and t) = 0.875 ---
**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\ **Post date:** [September 8, 2001, 8:03pm UTC](https://boards.straightdope.com/t/statistics-101/80421/3 "2001-09-08T20:03:56Z")
Or just think of it visually. The possibilities are HHH HHT HTH HTT THH THT TTH TTT So there are eight equally probable events, one of which has no heads. So the probability is you will get at least one head is 7/8.