sandy wrote:
A certain coin with heads on one side and tails on the other has a \(\frac{1}{2}\) probability of landing on heads. If the coin is flipped three times, what is the probability of flipping 2 tails and 1 head, in any order?
(A) \(\frac{1}{8}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{3}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{2}{3}\)
Easy way:Let's examine ONE case in which we get exactly 2 tails: HTT
P(HTT)=\(\frac{1}{2}\) *\(\frac{1}{2}\)*\(\frac{1}{2}\)=\(\frac{1}{8}\)
This, of course, is just ONE possible way to get exactly 2 TAILS.
Another possible outcome is THT
Here, P(THT)= (\(\frac{1}{2}\))(\(\frac{1}{2}\))(\(\frac{1}{2}\)) =\(\frac{1}{8}\)
As you might guess, each possible outcome will have the same probability (\(\frac{1}{8}\)). So, the question becomes "In how many different ways can we get exactly 2 TAILS ?"
In other words, in how many different ways can we arrange the letters HTT?\(\frac{3!}{2!}\)=3 WAY.
So P(exactly 2 Tails) = (\(\frac{1}{8}\))(3) = \(\frac{3}{8}\)