Question 1167500
<pre>
T T T T
T T T H
T T H T
T T H H   (1)

T H T T
T H T H   (2)
T H H T   (3)
T H H H

H T T T
H T T H   (4)
H T H T   (5)
H T H H

H H T T   (6)
H H T H
H H H T
H H H H

P(2 heads) = 6/16 = 3/8 = 0.375


Another way: there are four chances to get two heads, and there are 2^4 possible outcomes:   P(2 heads) = 4C2 / 2^4 = 6/16

And of course:  4 places to put the first H times 3 places for the 2nd H, but we divide by 2! because we've overcounted (the H's are indistinguishable) so we get  (4*3/2!)/2^4 = 6/16