If a coin is tossed a sequence of times, what is the probability that the first head will occur after the fifth toss, given that it has not occurred in the first two tosses?
(Assuming that this is a fair coin as it would be stated
otherwise.)
Since its given that it has not occurred in the first two tosses, we can
assume those two are tails and focus on the next four tosses.
We can then focus on the last four tosses because we need to find the
probability that the first head is after the fifth toss so it will have
to be if the first head is on the sixth toss: TTTH.
The probability of tails on the 3rd toss is \(\frac{1}{2}\)
The probability of tails on the 4th toss is \(\frac{1}{2}\)
The probability of tails on the 5th toss is \(\frac{1}{2}\)
The probability of heads on the 6th toss is \(\frac{1}{2}\)
So, we just combine these probabilities together to get: (\(\frac{1}{2}\))(\(\frac{1}{2}\))(\(\frac{1}{2}\))(\(\frac{1}{2}\)) = \(\frac{1}{16}\)
OR
Since its only four tosses to worry about, we can find all the
possible outcomes:
HHHH
THHH
TTHH
HTHH
TTTH
HHTH
HTTH
THTH
TTTT
HHHT
THHT
TTHT
THTT
HTHT
HHTT
HTTT
This gives us 16 outcomes but only one of them, TTTH, will be a success.