The easiest way to solve this problem is probably to just write down all the different possible outcomes of four consecutive tosses.
The possibilities are:
HHHH, HHHT, HHTH, HHTT,
HTHH, HTTH, HTHT, HTTT,
TTTT, TTTH, TTHT, TTHH,
THTT, THTH, THHT, THHH.
So we get 16 possible outcomes.
Answer: c. 16.
Comments
Leave a comment