Jane121393 wrote:
Hi
VeritasKarishma could you explain for me why my solution is wrong?
My approach: 2 ways for the contestant can't taste all 3 samples: He tastes 3 cups of 1 samples and 1 cup of another sample OR he tastes 2 cups of each 2 samples.
- 1st way: choose 1 sample out of 3 to provide 3 cups: 3C1. Then, 1 sample out of the remaining 2 to provide 1 cup => 2C1 * 3C1
=> 3C1 * 2C1 * 3C1 = 18
- 2nd way: choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2. Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> 3C1 * 3C2 * 2C1 * 3C2 = 54
=> total probability = (18+54)/ 9C4 = 4/7
I can't understand what's wrong with my approach. Please help. Thanks so much.
Hi
Jane121393For your 2nd way, you are actually duplicating the count. Lets take an example below:->
Assuming there are there three samples namely S1, S2, and S3. The tea blowing to each on them be S1T1, S1T2, S1T3, S2T1, etc.
S1T1 means Tea cup of 1st variety from sample 1. similarly others.
Now considering your 1st step for way 2 -
[b]choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2You choose sample S1 and then T1 and T2 cup, so you chose S1T1 and S1T2.
In your second step -
Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> - you choose sample S2 from remaining S2 and S3 and then 2 cups from it , assuming you choose S2T1 and S2T2.
So finally you have S1T1, S1T2, S2T1 and S2T2.
Now if you choose sample S2 in your 1st step and Cups 1 and 2 -b][b]choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2,
You will get S2T1 and S2T2
And if you choose sample S1 in your 2nd step and then cups 1 and 2 in your 2nd step - [b]Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2 - you will have S1T1 and S1T2.
So finally again you have S2T1 and S2T1 and S1T1 and S1T1. The only difference is that the order in which you selected is different. But here in selection, the order is irrelevant.
So effectively you are taking the same case twice.