I flip a coin 5 times in a row. The 32 possible outcomes of this
are listed below.
HHHHH | HHHHT | HHHTH | HHHTT | HHTHH | HHTHT | HHTTH | HHTTT |
HTHHH | HTHHT | HTHTH | HTHTT | HTTHH | HTTHT | HTTTH | HTTTT |
THHHH | THHHT | THHTH | THHTT | THTHH | THTHT | THTTH | TTHHH |
TTHHH | TTHHT | TTHTH | TTHTT | TTTHH | TTTHT | TTTTH | TTTTT |
Solution:
To do this, we can just count. In the above list, the possibilities with
exactly two heads are in bold, and there are 10 of them. There are 32
possibilites shown, so the answer is
fill
Solution:
Again, we can just count, but we have to notice WHAT we count. We are told
that the first flip came up tails, so we look only at the lines where the
outcomes start with a T (the bottom two lines). In those lines, there are 6
entries that have exactly two heads, out of 16 total entries. So, the
probablity is
If you prefer, you can use the conditional probability formula. In this
case, it says
Finally, another way to see the same result is this: since the flips are
independant, when I tell you that the first was a tail, we might as well
forget we did it at all. This changes the problem to ``what is the
probability of flipping a coin four times and getting exactly two
heads. There are six ways for this to happen (HHTT, HTHT, HTTH, THTH, THHT,
TTHH), and possible ways for the four coin flips to turn up.
Thus, again we get 6/16.
fill