The answer is: the probability of getting one and only one head across three tosses of a fair coin is .375.

Note,

• N=3
• N!=6
• X=1 (one head)
• X! = 1
• p=.5 (fair coin)
• (1-p)=.5
• (N-X)! = (3-1)! = 2! = 2

• Note that the total number of all possible combinations (including the outcome in which we're interested) is 2 X 2 X 2 -- 3 tosses, 2 possible outcomes for each toss. That is, the total number of possible outcomes is 8.
  
  
• Note that the total number of possible ways we can get one (and only one) head is 3 -- the combinations formula would
  give that to us by dividing
  
  N! (=6)
  by
  [X!(N-X)!] (=1! X 2! = 2)
  
  
• You then figure out how p (=.5) factors in to the end of the formula: p^x by (1-p)^n-x
  
  
• .5¹ multiplied by (1-.5)² = .125.
  
  
• So, the probability of rolling just one head in three tosses is 3 X .125=.375.
  
  
• Note you could also get that by looking at the combination tree
  
  
• You can do that in this example because it's a fair coin--the probability is .5. The ski slope example was a bit more complicated, because the probability of an event (breaking a leg) was not equal to the probability of a non-event (not breaking a leg).