SOLUTION: If a hedgehog crosses a certain road before 7.00 a.m., the probability of being run over is 1/10. After 7.00a.m., the corresponding probability is 3/4. The probability of the hedge

Click here to see ALL problems on Probability-and-statistics

Question 868940: If a hedgehog crosses a certain road before 7.00 a.m., the probability of being run over is 1/10. After 7.00a.m., the corresponding probability is 3/4. The probability of the hedgehog waking up early enough to cross before 7.00 a.m., is 4/5. What is the probability of the following events:
(a) the hedgehog waking up too late to reach the road before 7.00 a.m.
(b) the hedgehog waking up early and crossing the road in safety
(c) the hedgehog waking up late and crossing the road in safety
(d) the hedgehog waking up early and being run over
(e) the hedgehog crossing the road in safety.
PLEASE HELP!
Found 2 solutions by ewatrrr, rothauserc:
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi
Interesting :)
Before 7am: P(awake = 4/5 = .8), P(hit) = .10, P(hit after7) = .75
(a) the hedgehog waking up too late to reach the road before 7.00 a.m. 0r 20%
(b) the hedgehog waking up early and crossing the road in safety
(c) the hedgehog waking up late and crossing the road in safety
(d) the hedgehog waking up early and being run over
(e) the hedgehog crossing the road in safety.

Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
Note that a - d are mutually inclusive events and e is mutually exclusive events
a) 1 - 4/5 = 1/5 probability of waking up too late to reach the road before 7.00 a.m.
b) 4/5 * (1 - 1/10) = 36/50 = 18/25 probability of waking up early and crossing the road in safety
c) (1 - 4/5) * (1 - 3/4) = 1/20 probability of waking up late and crossing the road in safety
d) probability of waking up early and being run over = 4/5 * 1/10 = 4/50 = 2/25
e) probability of crossing the road in safety = 18/25 + 1/20 = 72/100 + 5/100 = 77/100