probability questions (1 Viewer)

[stupid idiot]

Suppose two balls are drawn with replacement from a jar containing 1 red ball, 1 white ball, and 1 black ball. If order is important, the possible samples are

{(B,W), (W,B), (B,R), (R,B), (W,R), (R,W), (B,B), (W,W), (R,R)};

and the total number of samples is 3² = 9. If order is not important, the possible samples are

{(B,W), (B,R), (W,R), (B,B), (W,W), (R,R)};

and the total number of samples is ^(3+2-1)C₂ = 6

Now, consider the probability of drawing 2 white balls. It first appears that if the sample space is ordered, this probability is 1/9, but if it is unordered, the probability is 1/6. What is wrong with this reasoning, and what is the correct probability?

I know the first way (the ordered way) is correct, but why? Can someone explain this to me please.

And can someone derive for me or explain to me with logics the formula: ^(n+r-1)C_r for unordered selection with replacement.

 

[Affinity]

imagine going to the supermarket, .. the fruit and vegetable section.
and imagine the fuits are lined up in the order of apples oranges and banana.. (how orignial.

suppose you want to buy ... say R fruits, and you ask yourself how many combinations can you get (assuming *cough* all fruits within one pile are identical)

you start with apples, you have 2 actions to choose from, wither (1) prick fruit, or (2) move to next pile.

together there are R picks and N-1 moves (where N is the number ofdifferent piles)

and depending on the order you execute these actions, you will always get a different pile.
and the question will be equivalent to one where you choose R out of N+R-1 actions to be 'picks' and there you go N+R-1 C R
hmm hope I am remotely understandable

 

[stupid idiot]

Yep i got ya. thanks.

Someone explain the first part of my question please

 

[Affinity]

first spot : 3 choices
second spot: 3 choices
are they independent? yes
so 3*3 = 9

 

[hainsay]

hmmm. shouldn't u already noe this?

the problem with the unordered pairs is that not all outcomes are equally likely, its like u expecting the bottom team to beat the top team. In a perfect probability world, with nine trials, u would get WW once, because each time, the chance of drawing a white is 1/3. u can get WB or BW to get a white and a blue. probability of getting BW is 1/3*1/3. probability of getting WB is 1/3*1/3, so probability of WB or BW is 2/9.

so the probability of each term in the unordered set is:

{WW, WB, WR, BB, BR, RR}
1/9, 2/9, 2/9, 1/9,2/9,1/9

with a total of 1