Binomial distribution - coin toss I would be really happy, if someone would explain how I solve this;
Let p denote the probability for “head” in a coin toss, p = 0.5. The coin is thrown N times. What is the probability of the following events?
*: Please provide details, not just the solution.
N = 3. First head, then head, then tail:
I know the binomial formula but cannot figure out how i solve this. 
Regards Ida 
 A: The event you are asking about can be seen as the combination of three independent events: head (probability 0.5), head (probability 0.5), tail (probability 0.5). Tossing a coin three times or tossing three (numbered: 1st, 2nd and 3rd) coins are equivalent events.
The joint probability for independent events is the product of the probabilities of each single event (see for example here), so the joint probability of your event is 
0.5 * 0.5 * 0.5 = 0.125
A: You are tossing 3 times. It means there is total 2^3 = 8 possible outcomes and each outcome have the same probability 1/8 = 0.125. You have chosen 1 of the 8 outcomes and thus it has probability 1/8 = 0.125
This means the probability for your combination of h,h,t is just as likely as h,h,h or t,t,t or in any other of the 8 combinations. It seems a bit counter intuitive to humans, because they think (h,h,t) is more likely than (h,h,h) and that intuition is also correct, if the order does not matter.  
2h and 1t in any order have 3 outcomes, which you can calculate as:
3!/(2h! *1t!)= 6/2= 3 possible combinations/outcomes. This probability is 3 out of 8 possible outcomes or 3/8= 0.375. But if you pin point a unique outcome (h,h,t) from those 3, it will have the probability 3/8 / 3 = 1/8 
