Answer:
(i)

(ii)

(iii)

(iv)

Explanation:
Let
be the event that the dog has gone home,
be the event that the dog has gone to the picnic area, and
be the event that the dog has gone to the park.
Assuming all the three events are equally likely to happen, so,

Now, let
and
bet the event of found and lost of the dog.
Given that the chances of finding the dos are
and
, if the dog is in the picnic area and woods respectively.i.e
,
and


(i) The probability that the dog will be found in the park
Probability of going the dog to park
Probability of found in the park
[using equations (1) and (2)]


(ii) If the dog is in home, the chance of finding the dog is 100%.
So, the probability that the dog will be found at home
The probability that the dog has gone home

[ from equation (1)]
(iii) Given that the dog is found in the park, so, the probability of founding the dog in the picnic area of the park

[using equations (1), (2) and (4)]

(iv) Given that the dog is lost, so, the probability of losing the dog in the woods

[using equations (1), (3) and (5)]
.