Probability Distribution-Without Replacement

[kia68]

I am stuck on the following Question. I worked out the answer for replacement but I am completely lost on the "without replacement" aspect. Any help or explaination would be greatly appreciated

A bag contains five tokens numbered 2, 3, 6, 7, and 8.
Two tokens are taken in succession out of the bag without replacement.
a) Create the probability distribution for “x” being the number of odd numbered
tokens drawn.

 

[Steven G]

What values can x take on? Remember that x is the number of odd numbered token picked.

 

[kia68]

I'm not sure simply because the question says 2 tokens are taken in succession. That's the part that confuses me because I could take 2 and 3 out of the bag at the same time but what about the possibilities of taking out 2 and 7? However. because there are no replacements, once I take 2 out of the bag I can have any other combinations of 2 and another number

 

[Dr.Peterson]

I'm not sure simply because the question says 2 tokens are taken in succession. That's the part that confuses me because I could take 2 and 3 out of the bag at the same time but what about the possibilities of taking out 2 and 7? However. because there are no replacements, once I take 2 out of the bag I can have any other combinations of 2 and another number

The numbers will be small enough that you can just list all the possibilities, and count the number of odd tokens in each case:

2,3​

2,6​

2,7​

2,8​

3,2​

3,6​

...​

After you've done that, and have a better sense of how it works, you may, if you wish, look for a more "mathematical" way to count the number of ways to get x odd tokens.

But surely you can answer Jomo's question! What is the most odd tokens you can get? What is the fewest?

 

[kia68]

This is what I’ve deducted and then placed them into a probability distribution table. But I’m assuming this is with replacement because each number shows up more than once

 

[Dr.Peterson]

This is what I’ve deducted and then placed them into a probability distribution table. But I’m assuming this is with replacement because each number shows up more than once

Looks good to me.

Why do you think this is "with replacement"? Each possibility in your list consists of two different numbers, so you clearly didn't replace anything. With replacement would have had (2,2), (3,3), and so on also in the list.

The one (trivial) thing I would have done differently is to initially write the unsimplified probabilities (6/20, 12/20, and 2/20) in the table to make it easy to check that they add up to 1, and then simplify.

 

[kia68]

ooo ok. My understanding was that if I removed a token from the bag I wouldn't be able to use it again. So if I took out 2 and 3, then I couldn't have 3 and 2. I'll unsimplify the probabilities. Thank you soo much

 

[Dr.Peterson]

Right. Each possibility in the list is separate from the others, so we are starting over each time. The non-replacement is only within the process of creating any one possibility (that is, one outcome in the sample space).