A fair coin is tossed 4 times. What is the probability that two...

Total number of possible outcomes when a fair coin is tossed 4 times is $2^4 = 16$.

Let $E$ be the event that two heads do not occur consecutively. The favorable outcomes are the sequences of length 4 that do not contain 'HH'.

We can count these by the number of heads in the sequence:
Case 1: $0$ heads
TTTT ($1$ outcome)

Case 2: $1$ head
HTTT, THTT, TTHT, TTTH ($4$ outcomes)

Case 3: $2$ heads
To avoid consecutive heads, the heads must be separated by tails. The possible arrangements are HTHT, HTTH, THTH ($3$ outcomes)

Case 4: $3$ or $4$ heads
It is impossible to place $3$ or $4$ heads in $4$ tosses without at least two heads being consecutive. ($0$ outcomes)

Total number of favorable outcomes = $1 + 4 + 3 = 8$.

Required probability = $\dfrac{8}{16} = \dfrac{1}{2}$.

Answer: $1/2$