Whats the answer please
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
Let us record the station number 1, 2 or 3 for each family member A, B or C.
I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.
From table it is clear that the total outcomes possible are 27.
We know that, probability can be defined as,
a) All Members Assigned to the Same Station.
Cases for all members being assigned to same station are as follows:
[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).
Therefore,
b) At Most Two Members Assigned to the Same Station.
It means that maximum of 2 members can have the same station. Cases for this situation are as follows:
[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],
[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]
(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).
Therefore,
c) All Members Assigned to a Different Station.
For this scenario, we have the following results:
[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).
Therefore,
- 5x-3y=11---(1)
- 3x+4y=4---(2)
(1)×3 and (2)×5
- 15x-9y=33-(3)
- 15x+20y=20-(4)
Subtract both
- Put in eq(1)