F.5 Maths M1 probability

(a) Let 'E' represent a step to the east, 'N' represent a step to the north. For example, if Andy walks eastward firstly for two units, then a step to the north (that's where he meets junction B), then a step to the east, finally two steps to the north to arrive point C. Then his route is represented as 'EENENN'.

Therefore the number of routes he can choose equals to the number of ways to fill in 3 'E's and 'N's in six places, which equals to 6C3 = 20.

Hence the answer in (a) is 20.

(b) (i)

He can only walk to the north or walk to the east. Therefore P(he goes to east) = 0.7. Consider the probability of a few cases to happen:

P('EENNEN') = 0.7 × 0.7 × 0.3 × 0.3 × 0.7 × 0.3 = (0.3³)(0.7³)
P('ENENNE') = 0.7 × 0.3 × 0.7 × 0.3 × 0.3 × 0.7 = (0.3³)(0.7³)
P('NNNEEE') = 0.3 × 0.3 × 0.3 × 0.7 × 0.7 × 0.7 = (0.3³)(0.7³)

You may notice the probability of every case to happen is the same.
Therefore P(Andy will arrive at C) = 20 × (0.3³)(0.7³) = 0.18522

(ii)
If Andy's first three steps are 'EEN', 'ENE' or 'NEE', he will pass junction B.
No. of cases of passing jucntion B = 3 × 3C1 = 9
⇒ P(arrive at C without passing junction B) = 9 × (0.3³)(0.7³) = 0.083349