If the arrival of major road vehicles can be described by the poisson distribution, and the peak hour volume is 1000 veh/hr. the value of the critical gap is 4 seconds. the expected number of available gaps during the peak hour is equal to:

Expected no. of vehicles per hour = 1000
expected no. of vehicles per 4 minute interval = 1000/(3600/4) = 10/9 = λ

Using the poisson distribution, probability that no vehicles within the 4 second interval
P(0)= λ ^0 e^(- λ )/ 0!
= (10/9)^0 e^(-10/9) / 1
= e^(-10/9)
= 0.3292 (approx.)

In an hour, there are n=3600/4=900 such intervals, each with probability of p=0.3292 occurring.

Applying binomial distribution,
expected number of gaps
=np
=900*0.3292
= 296.3

the expected number of gaps is 296 (to the nearest unit).

Note: in fact, the actual expected number will be higher because gaps do not line up at 4 second intervals.  If a vehicle passes at the 1 second of the interval, a gap can commence at the next second.  Therefore the expected value could/should be slightly higher.

Logical equivalent of the give question is its contrapositive