Answer:
a) P(C/L) = 0.6668
b) P(C/L') = 0.1875
Step-by-step explanation:
Let's call C the event that Brian takes Commuter North, B the event that Brian takes Bursley Baits, N the event that Brian takes Northwood Express, L the event that Brian is late and L' the event that Brian is not late.
First, there is equal probability of taking any given bus so, P(C)=P(B)=P(N)=1/3
Now, the probability P(C/L') that Brian took a Commuter North bus given that he is late is calculated as:
P(C/L) = P(C∩L)/P(L)
Where P(L) = P(C∩L) + P(B∩L) + P(N∩L)
Then, the probability P(C∩L) that Brian takes a Commuter North bus and it is late is calculated as:
P(C∩L)= (1/3)*(0.5) = 0.1667
Because, there is a probability of 1/3 to takes Commuter North Bus and if Brian takes Commuter North there is a 0.5 chance to be late.
At the same way, we get:
P(B∩L) = (1/3)(0.2) = 0.0667
P(N∩L) = (1/3)(0.05) = 0.0167
So, P(L) and P(C/L) are equal to:
P(L) = 0.1667 + 0.0667 + 0.0167 = 0.25
P(C/L) = 0.1667/0.25 = 0.6668
For part b, the probabilities of C, B and N changes and are equal to:
P(C) = 0.3
P(B) = 0.1
P(N) = 0.6
Then, the probability P(C/L') that he took a Commuter North bus given that Brian was not late to class is calculated as:
P(C/L') = P(C∩L')/P(L')
Where P(L') = P(C∩L') + P(B∩L') + P(N∩L')
So, P(C∩L'), P(B∩L') and P(N∩L') are equal to:
P(C∩L') = 0.3*(0.5) = 0.15
P(B∩L') = 0.1*(0.8) = 0.08
P(N∩L') = 0.6*(0.95) = 0.57
It means that P(L') and P(C/L') are equal to:
P(L') = 0.15 + 0.08 + 0.57 = 0.8
P(C/L') = 0.15/0.8 = 0.1875
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