Hi,
Can someone confirm why we didn’t arrange these combinations in different positions.
I thought that Jane here in both the cases (Junior, Senior, Jane and Senior, Senior, Jane) could be rearranged in 3 manners making total 32 happy cases ([Jane(1) Junior(2) Senior(3)]3 and [Jane(1) Senior(2) Senior(3)]3)
If we similarly calculate the denominator, we will have 108 combinations (JJS- 3232 + JSS- 3322]
Let’s assume that the three juniors are Jane (J1), J2, and J3, and the three seniors are S1, S2, and S3. Given that at least one junior and at least one senior are selected, there are only two possible combinations: two juniors and one senior or two seniors and one junior
As you can see from the list above there are a total of 18 possible combinations given that at least one junior and at least one senior are selected.
Our happy case are the one’s with J1(Jane) in it (which are 9 in total); thus , proab = \dfrac{9}{18}=\dfrac{1}{2}
Hi, can you help with a combinations approach? I am getting a different answer.
can you provide the rough work ?
There are a total of C(6,3) = 20 ways to select 3 students out of 6. Since at least one junior and at least one senior must be selected, we can subtract the number of ways to select only juniors or only seniors from the total number of ways. There are C(3,3) = 1 way to select only juniors and C(3,3) = 1 way to select only seniors. So there are 20 - 1 - 1 = 18 ways to select at least one junior and at least one senior.
Now let’s find the number of ways to select Jane and at least one senior. Since Jane is already selected, we need to select 2 more students from the remaining 5 (2 juniors and 3 seniors). There are C(5,2) = 10 ways to do this. However, this includes the case where both students selected are juniors. There is C(2,2) = 1 way to do this. So there are 10 - 1 = 9 ways to select Jane and at least one senior.
Therefore, the probability that Jane is selected given that at least one junior and at least one senior are selected is 9/18 = 1/2.
