Hi! I've got the task. That must be tricky and fraught with pitfalls. But I've solved that too easy and fast, and now I concern if my solution is right? May be I missed something important?
The case:
There are 3 ferry stations (A, B, and C), there is 1 ferry at each station. The destination of A's ferry is either station B or station C, the destination of B's ferry is either station A or station C, the destination of C's ferry is either station A or station B.
The destination is chosen right before departure, for each of two possible stations the probability of choosing = 0.5. The choice of station doesn't depend on choices of the others.
Each ferry operates twice weekly, on Monday and Friday. The departure station on Friday is the same as arrival station on Monday.
Again, the arrival station on Friday is chosen with 0.5 probability.
I need to find the probability that after Friday's trips there will be 2 ferries at some station.
My solution is:
1) Sample space for Monday trip:
Probability of any event = 1/8
2) Sample space for Friday trip:
In case of BAA option at the first step:
In case of BCA option at the first step:
We may notice that if there were 2 ferries at some station on the first step, the number of suitable events = 5. If not, the number of suitable events = 6.
So, the answer is: 1/8*(5/8 + 6/8 + 5/8 + 5/8 + 5/8 + 5/8 + 5/8 + 6/8 ) = 42/64 = 0.65625
Is that correct? Or I need to use some fancy formula like conditional probability or something else?
The case:
There are 3 ferry stations (A, B, and C), there is 1 ferry at each station. The destination of A's ferry is either station B or station C, the destination of B's ferry is either station A or station C, the destination of C's ferry is either station A or station B.
The destination is chosen right before departure, for each of two possible stations the probability of choosing = 0.5. The choice of station doesn't depend on choices of the others.
Each ferry operates twice weekly, on Monday and Friday. The departure station on Friday is the same as arrival station on Monday.
Again, the arrival station on Friday is chosen with 0.5 probability.
I need to find the probability that after Friday's trips there will be 2 ferries at some station.
My solution is:
1) Sample space for Monday trip:
- BAA
- BCA
- BCB
- BAB
- CAA
- CCA
- CCB
- CAB
Probability of any event = 1/8
2) Sample space for Friday trip:
In case of BAA option at the first step:
- ABB
- ABC
- ACB
- ACC
- CBB
- CBC
- CCB
- CCC
In case of BCA option at the first step:
- ABB
- ABC
- AAB
- AAC
- CBB
- CBC
- CAB
- CAC
We may notice that if there were 2 ferries at some station on the first step, the number of suitable events = 5. If not, the number of suitable events = 6.
So, the answer is: 1/8*(5/8 + 6/8 + 5/8 + 5/8 + 5/8 + 5/8 + 5/8 + 6/8 ) = 42/64 = 0.65625
Is that correct? Or I need to use some fancy formula like conditional probability or something else?