altairahmad
I tried to attempt this question with fundamental counting principle. Can you please point out my mistake here.
No of ways of 2 cups of same type and 2 of different types AABB = 9(first type)x2(should be same as first)x6(second type)x2(should be same as second) = 216
Different ways of (AABB)= 4!/(2!x2!) = 6
AABB = 216/6 = 36
No of ways of # cups of same type and 1 of different type AAAB = 9(first type)x2(should be same as first)x1(should be same as first)x6(second type) = 108
Different ways of (AAAB) = 4!/3!
AAAB = 27
Total ways = (9x8x7x6)/4! = 126
Probability = (27+36)/126=1/2
I know that I am doing a conceptual mistake but I can't figure out where.
There are two separate issues with this solution. You first count how many ways to get the specific sequence AABB. But to get two teas each of two types, we could have various orders: ABAB, say, or BAAB. As you correctly calculate, there are 4!/(2!*2!) orders we can put these four letters in. So to find all of the sequences of A, A, B, B, you need to
multiply by 4!/(2!)(2!), and not divide by it.
But then there is a second issue, which is quite subtle, that only affects the analysis of the first case. If you assume the first tea can be anything, the second must match, and then the final teas must match each other, you'll be counting the possibilities of getting AABB and also of getting BBAA. But then if you count all the orders you can put A, A, B, B in, you'll be rearranging AABB to get BBAA, and you'll thus be counting BBAA for a second time. So doing this double-counts every possibility, and we need to divide by 2 to get the right answer. That double-counting issue is the reason many solutions earlier in this thread proceeded by first selecting the two types of tea (3C2, or 3 ways). Then you get the sequence AABB in 3C2*(3)(2)(3)(2) ways, with no double-counting.
Finally, if you're solving this way, you're assuming you're selecting the four teas in order, so the denominator needs to be 9*8*7*6, and not 9C4. Putting all of that together, the answer becomes:
[ ((9)(2)(6)(2)/2) * 4!/(2!*2!) + (9)(2)(1)(6) * 4 ] / (9)(8)(7)(6) = (9*2*6*6 + 9*2*6*4)/(9)(8)(7)(6) = (12 + 8)/8*7 = 20/56 = 5/14
I think it's tricky to get to the right answer using this precise method, so I'd definitely prefer a different approach here.
Anant87
Can someone please explain to me what is wrong with this approach ?
I am calculating the probability of the opposite event - we need to make sure we select atleast 1 cup from each sample -
Options for selecting a cup from the first sample - 3C1
Options for selecting a cup from the second sample - 3C1
Options for selecting a cup from the third sample - 3C1
Options for selecting the fourth cup - 6C1 (as there are 6 options remaining in total)
So req probability = 1- (3C1 * 3C1 * 3C1 * 6C1)/9C4........which is -ve so obviously wrong
There are a couple of issues here. First, there are many ways to select one cup from each type of tea. Maybe the first two cups both are the same type of tea, and the final two are different. Or maybe the first cup comes from the third type of tea, not the first. You haven't accounted for any of those possibilities. There are solutions earlier in this thread that correctly count how many ways to pick one cup of each type, which you may want to look at. Just as importantly, when you're counting the ways to get one cup of each type, you're counting as if you're selecting cups in order. That's perfectly fine, but if you do that in the numerator of a probability, you also need to count the total number of possibilities (the denominator) as if you're selecting in order. So if you use this approach, your denominator would need to be 9*8*7*6, and not 9C4.
AndrewN
I then used, more or less, a slot-based approach to select valid options at each step.
__ __ __ __
The first cup can be selected from any cup, so the probability is 9/9. For the sake of simplicity, I simulated the first A being chosen:
AAA
BBB
CCC
The second cup can also be selected from any of the remaining cups, so the probability is 8/8. Again, I simulated an A being chosen, with an eye on exhausting the A's:
AAA
BBB
CCC
The third cup is where you have to be careful. You might think that of the remaining 7 cups, only 4 can be chosen: the third A + any of the three B's or C's, but this ignores the fact that the non-A cup is not yet determined in this scenario, so there can actually be 5 valid selections for the third cup: the third A + any of the three B's or C's + 1 of either B or C (to lock in a track).
I don't understand any of this solution once it gets to the third cup (it frankly doesn't make any sense at all to me -- if your first two cups are AA, the third cup can literally be anything). I think it's just coincidence that you've arrived at the right answer. Conceptually, the instant a solution assumes the first two teas are both "A", it cannot possibly be correct. That's because when the first two teas are the same (both A), it becomes less likely you'll try all three types of tea (compared to the situation when the first two teas are different). Since the probability you will get all three types of tea is different when you pick AA than when you pick AB/AC, it becomes mandatory to divide the problem into cases at that point.
You can adjust that solution. If we first pick any tea at all, call it 'A', then:
• 2/8 of the time we pick A on the second selection, and 1/7 of the time we pick A with our third selection. Then the fourth selection doesn't matter, so we get a sequence like AAAX (2/8)(1/7) of the time
• 2/8 of the time we pick A on the second selection, and 6/7 of the time we pick something else with our third selection. Then the fourth selection must either be 'A' or must match the third, so there are 3 things we can pick out of 6 possibilities, and the probability we get a sequence starting with AAX (where X is not A) is (2/8)(6/7)(3/6), if we don't want all three types of tea
• 6/8 of the time we pick something other than A on the second selection. Then our third and fourth selections must either be A, or must match our second selection. So we have 4 options with our third selection, and 3 for our last selection, and we get a sequence starting AX (where X is not A) that does not have all three types of tea (6/8)(4/7)(3/6) of the time
Those are all the possible cases, and adding them, we get
(2/8)(1/7) + (2/8)(6/7)(3/6) + (6/8)(4/7)(3/6) = (1/4)(1/7 + 3/7) + (3/4)(4/7)(1/2) = 1/7 + 3/14 = 5/14
But, as I said in an earlier post, the method I think is best for this problem is the one mira93 used: it's easy to work out the probability you don't get tea A on any selection (that's just (6/9)(5/8)(4/7)(3/6)), and then we multiply by 3 to get the answer, because we want to count the possibilities where we avoid tea B and tea C also.