The figures haven't been displayed so we'll use algebraic symbols for the probabilities.
w is the probability of the hedgehog waking before 7am; 1-w is the probability of it waking after 7am.
r is the probability of it being run over before 7am; 1-r is the probability of it not being run over before 7am.
s is the probability of it being run over after 7am; 1-s is the probability of it not being run over after 7am.
Below is a diagram of possible outcomes for the hedgehog in a typical day. There appear to be 5 outcomes (top to bottom):
the hedgehog wakes up:
(1) before 7am, crosses the road, but (later crosses the road to return home and) gets run over after 7am.
(2) before 7am and crosses the road safely (each time).
(3) before 7am, crosses the road and is run over and killed.
(4) after 7am, crosses the road and is run over and killed.
(5) after 7am and crosses the road safely (each time).
An alternative to this diagram is to omit (1) and (2), and replace by (1) labelled w(1-r) as the safe outcome. Another alternative (not shown) is a safe outward journey but two possible outcomes for a return journey. We would then have 6 possible outcomes. The question does not clarify how many times the road is to be crossed.
The probability of each outcome is shown algebraically. The sum of the outcomes=1 as it should be:
sw-rsw+w-rw-sw+rsw+rw+s-sw+1-s-w+sw=1.
The probability of remaining safe all day=w(1-r)(1-s)+(1-w)(1-s)=w-rw-sw+rsw+1-w-s+sw=-rw+rsw+1-s=(1-s)(1-rw).
EXAMPLE using w=80% or 0.8; r=10% or 0.1; s=20% or 0.2:
(1)=(0.8)(0.9)(0.2)=0.144 or 14.4%;
(2)=(0.8)(0.9)(0.8)=0.576 or 57.6%;
(3)=(0.8)(0.1)=0.08 or 8%;
(4)=(0.2)(0.2)=0.04 or 4%;
(5)=(0.2)(0.8)=0.16 or 16%.
Total outcomes=1 or 100%.
Total safety: 0.576+0.16=0.736 or 73.6%.
(1) and (2) can be replaced by w(1-r).
To use the diagram to compute probabilities, follow each track to an outcome, multiplying the probabilities encountered along the way.
(a) 1-w=0.2 or 20%;
(b)=(2)=0.576 or 57.6% (hedgehog crosses the road twice); alternatively replaced by w(1-r)=(0.8)(0.9)=0.72 or 72%;
(c)=(5)=0.16 or 16%;
(d)=(3)=0.08 or 8%;
(e)=(2)+(5)=0.736 or 73.6%; or, alternatively, 0.72+0.16=0.88 or 88%, depending on whether the hedgehog crosses the road more than once (that is, on the outward and inward (return home) journeys).
The diagram, solution and examples are only for the purposes of illustration of a method that is easy to follow.