Number of possible outcomes where Head is recorded for a coin toss

[tmt1]

A coin is tossed 4 times.

Is there a way to determine mathematically what is the probability that exactly 2 heads occur?

By drawing a decision tree I can determine that it is 6/16, but this seems like an arduous process for larger numbers.

 

[MarkFL]

Using the binomial probability formula, we find:

\(\displaystyle P(X)={4 \choose 2}\left(\frac{1}{2}\right)^2\cdot\left(\frac{1}{2}\right)^{4-2}=\frac{6}{16}=\frac{3}{8}\)

 

[Greg]

Writing 'em out helps but I wouldn't want to attempt that on a "large" sample space.

There are 16 possible outcomes and 6 possibilities where there are exactly two heads. 6/16 = 3/8.

 

[HallsofIvy]

Or: one possible outcome for 'two heads in four tosses" is HHTT. The probability the coin comes up heads or tails on each toss is 1/2 so the probability of that is (1/2)^4= 1/16.

But there are $$\frac{4!}{2!2!}= \frac{4(3)(2)(1)}{(2(1))(2(1))}= \frac{4(3)}{2}= 6$$ different possible orders (they are "HHTT", "HTHT", "HTTH", "THTH", "THHT", and "TTHH" but you don't have to write them out to know there are 6) so that the probability of "two heads in four coin tosses" is $$\frac{6}{16}= \frac{3}{8}$$