You flip a true coin 9 times. What is the probability of getting heads 4 times and tails 4 times??? You can get tails and heads in any order. You don't have to get 4 consecutive heads in a row then 4 consecutive tails in a row. For example, you could flip, heads;tails;heads;tails;heads;tails;heads;tails
0. That ninth coin toss makes it impossible to get both heads and tails exactly four times each. Or do you mean heads for times or more, and tails four times or more?
If you mean minimum four heads and four tails in 9 tosses, so, either 5 heads and 4 tails, or 5 tails and 4 heads. The answer is 0,492. If butterfingers, and 9 was a typo, and you meant 8 tosses, the answer is 0,273.
Another one, a little simpler: If you toss a coin three times, what's the probability of getting one tails and two heads?
I love probability problems. They often end up being much harder to think about than people first realize. I recently read The Drunkards Walk: How Randomness Rules Our Lives. It was a great read.
Yeah, love them too. This one is a bit harder than the first one, if he meant minimum four of each, instead of exactly four. Hope the next one gets harder though.
Wouldn't the answer be 12.5% Wouldn't the answer be 12.5% for every combination of heads and tails you would chose??? for example: Heads: heads; heads = 12.5% Tails : Heads : Tails 12.5% ???
Nope. You have 2^3 outcomes with three coin tosses. Since this is so few tosses, it's possible to write them all down: H-H-H H-H-T H-T-H T-H-H H-T-T T-H-T T-T-H T-T-T Eight outcomes. Of those eight outcomes, three are favorable, T-H-H, H-T-H, and H-H-T. So, 3/8=0,375 If there's too many coin tosses, and you don't bother to write down all the possible and favorable outcomes, just use this simple formula: ((A+B)!)/((A!)*(B!)*(2^(A+B))) A=the face you want to find the probability for, in this case we want one tails, so it's 1. B=the other face, in this case, heads, so it's 2. We know it's two, because there's three coin tosses, and 3-1=2. != for those unfamiliar with it, means factorial, quick explanation: 4!=4*3*2*1 /=division *=multiplication ^=(I think you anglos call it power) basically 4^3=4*4*4 I sure hope i didn't mess up with those parentheses. And I sure hope people didn't start writing down all the possible outcomes for your original question, although that may explain why there hasn't been any responses.
That's also a way to start off. Still working with the modified OP post, that is at least four of each. By doing (1/2)^9 you find two things. 2^9=512. So that's all the different variations these 9 coin tosses can be. And 1/512 is the chance for each of those variations. Now you only need to find the favorable variations, which you find by 9!/(5!*4!). Which gives us 126. So, double that, since you can have either four heads and five tails, or five heads and four tails. And the calculation in this case is identical. So, 126*2=252. So, 512 possible variations, and 252 favorable. 252/512=0,492
