Probability without replacement

S

soggles

New member
Joined
May 11, 2020
Messages
10
Hello there,
I am working on this problem and cant figure out if i am on the right track or not.

A game involves picking up to 4 tiles out of a bag that contains 20 tiles. The bag contains 4 tiles with the number 1 printed on them, 6 with 2 on them, 3 with 3 on them and 7 with the number 4 printed on them.
You pick the tiles one at at time. If the 1st tile you choose has a 1 on it then you win. If the 2nd tile has a 2 then you win and similarly for the remaining choices. After each choice, the tile you picked isn't returned to the bag.
What is the probability of winning this game?

So far for working I have started to do a tree diagram but then noticed it started to get messy. Cause there is 20 tiles I have 20C4 for tile 1, 20C6, 20C3, 20C7

Thank you!
 
What has 20C4 to do with anything.

P(Winning on the first draw) = 4/20. If you win, do you keep drawing or does the game end?
 
soggles said:
A game involves picking up to 4 tiles out of a bag that contains 20 tiles. The bag contains 4 tiles with the number 1 printed on them, 6 with 2 on them, 3 with 3 on them and 7 with the number 4 printed on them.
You pick the tiles one at at time. If the 1st tile you choose has a 1 on it then you win. If the 2nd tile has a 2 then you win and similarly for the remaining choices. After each choice, the tile you picked isn't returned to the bag.
What is the probability of winning this game?

So far for working I have started to do a tree diagram but then noticed it started to get messy. Cause there is 20 tiles I have 20C4 for tile 1, 20C6, 20C3, 20C7
Click to expand...
It will help if you show us the tree as far as you have taken it, so we can see if you are making the wrong kind of tree, or filling it in incorrectly, or just need more patience.
 
win on 1st draw = 1
win on 2nd draw 1, 2
win on 3rd draw 1, 2, 3
win on 4th draw 1, 2, 3, 4

Does this help?
 
soggles said:
A game involves picking up to 4 tiles out of a bag that contains 20 tiles. The bag contains 4 tiles with the number 1 printed on them, 6 with 2 on them, 3 with 3 on them and 7 with the number 4 printed on them.
You pick the tiles one at at time. If the 1st tile you choose has a 1 on it then you win. If the 2nd tile has a 2 then you win and similarly for the remaining choices. After each choice, the tile you picked isn't returned to the bag. What is the probability of winning this game?
Click to expand...
If one wins it must be on the first, second, third or fourth pull.
On the first: \(\dfrac{4}{20}\)
On the second: \(\dfrac{16}{20}\cdot\dfrac{6}{19}\) That is Not on the first but on second.
So what are the other two possibles?
Then add the four together.
 
But the first draw could be a 2, 3, or 4, and the probability of a 2 on the second depends on that.
 
I realize my mistake during my rest period. To win on the second pull it could be:
\(2,2;~3,2;\text{ or }4,2\)
Then a win on the third could be:
\(2,1,3;~2,3,3;~2,4,3;~3,3,3;~4,1,3;~4,3,3;~4,4,3\) Did I miss any?
On the fourth is a nightmare.
 
Missed 313: 243:

I'll let the OP figure out the how to win on the 4th try.
 
Jomo said:
Meant to say you accidentally missed 343
Click to expand...
Thank you again. My frustration is not being able to see a pattern. what about the \(4's\)?
I have:\(4114,4124,4144,4314,4324,4344,4404,4424,4444\)
Please do check my list.
 
Here is how I do it, by brute forth. I only list three numbers since the 4th number is 4. Then I cross out the bad ones

111, 112 113, ---- all no good as they all start with 1

211 212 213 214 221 222 223 224 231 232 233 234 241 242 243 244

311 312 313 314 321 322 323 324 331 332 333 334 341 342 343 344

411 412 413 414 421 422 423 424 431 432 433 434 441 442 443 444

Notice how a whole column cancels out!
27 cases remain. Yuck!

So I get 2114, 2124, 2144, 2314, 2324, 2344, 2414, 2424, 2444, 3114, 3124, 3144, 3314, 3324, 3344, 3414, 3424, 3444, 4114, 4124, 4144, 4314, 4324, 4344, 4414, 4424 and 4444

For some reason your list only have 4's for the 1st position. Is that a mistake or am I missing something (making this list is hurting my brain). Your list with 4's first matches mine with a slight typo--you had a 0 instead of a 1
 
Jomo said:
Here is how I do it, by brute forth. I only list three numbers since the 4th number is 4. Then I cross out the bad ones

111, 112 113, ---- all no good as they all start with 1

211 212 213 214 221 222 223 224 231 232 233 234 241 242 243 244

311 312 313 314 321 322 323 324 331 332 333 334 341 342 343 344

411 412 413 414 421 422 423 424 431 432 433 434 441 442 443 444

Notice how a whole column cancels out!
27 cases remain. Yuck!

So I get 2114, 2124, 2144, 2314, 2324, 2344, 2414, 2424, 2444, 3114, 3124, 3144, 3314, 3324, 3344, 3414, 3424, 3444, 4114, 4124, 4144, 4314, 4324, 4344, 4414, 4424 and 4444

For some reason your list only have 4's for the 1st position. Is that a mistake or am I missing something (making this list is hurting my brain). Your list with 4's first matches mine with a slight typo--you had a 0 instead of a 1
Click to expand...
So how would i go about working it out from here?
 
soggles said:
So how would i go about working it out from here?
Click to expand...
As you can see I completely changed my ideas. I now see that there is no way to avoid the consideration of all possible cases.
That is due to the non-replacement. There are three ways to win on the second draw:
\(22,~32,~\&~42\) so \(\mathcal{P}(22)=\dfrac{6}{20}\cdot \dfrac{5}{19};~\mathcal{P}(32)=\dfrac{3}{20}\cdot \dfrac{6}{19};~\mathcal{P}(42)=\dfrac{7}{20}\cdot \dfrac{6}{19}\).
Here is an example of winning on the third draw; \(313\) is \(\mathcal{P}(313)=\dfrac{3}{20}\cdot \dfrac{4}{19}\cdot \dfrac{2}{18}\)
and an example of winning on the forth draw; \(2414\) is \(\mathcal{P}(2414)=\dfrac{6}{20}\cdot \dfrac{7}{19}\cdot \dfrac{4}{18}\cdot \dfrac{6}{17}\)
Of course, the down-side is the necessity of listing all the cases. But the upside is it clearly tracts what tabs have been removed and not replaced.
 
soggles said:
So how would i go about working it out from here?
Click to expand...
You find the probability of each possible pair listed.

Foe example to win on the 4th trial you would have to calculate p(4314) where 4314 means 1st you pick a tile with 4 on it, then a tile with a 3 on it, then a tile with 1 on it and finally a tile with a 4 on it. You have many many of these probabilities to compute. I certainly would not want to have to do this problem
 
You must log in or register to reply here.
Top