Probability of Drawing Objects From a Golf Bag

[lacrotix]

Homework Statement

This question is suppose to be solved at the Math 12 Principles level:

Suppose there are 5 white golf balls, 3 yellow golf balls, and 4 orange golf balls in a pocket of a golf bag. Three balls are randomly drawn, and not replaced...

What is the possibility of drawing 2 white golf balls and 1 yellow golf ball?

Homework Equations

A couple of equations, but my teacher strongly suggested the class to use logic and "inner gut reasoning" to solve a good portion of probability questions. He said that this would be more beneficial to us than trying to simply use a formula.

The Attempt at a Solution

I did:
\frac{5}{12} \:*\: \frac{4}{11} \:*\: \frac{3}{10} \:=\: \frac{1}{22} In which:

1. The (5/12) is suppose to represent the first draw being a white ball.
2. The (4/11) is suppose to represent the second draw being another white ball.
3. The (3/10) is suppose to represent the third draw being a yellow ball.
4. The (1/22) is suppose to represent the final calculated answer of the probability of the given problem.

However, the answer is suppose to be (3/22), which is my answer multiplied by 3. I feel I am almost close to reaching this solution, but I can't yet rationalize why I would need to multiply my answer by 3.

Can anybody help and clarify and explain this to me.

Thanks!

 

[Dick]

That would be correct if the problem said what is the probability of drawing the first ball white, the second ball white and the third ball yellow. The problem doesn't specify the order of drawing the balls. The first or second ball could also have been yellow.

 

[lacrotix]

So why would I need to multiply my answer by 3 then? I am confused about that.

 

[phinds]

So why would I need to multiply my answer by 3 then? I am confused about that.

You have calculated the probability of producing one of more than one mutually exclusive result sequences. What are the possible sequences

 

[HallsofIvy]

There are n! (n factorial) ways to order n objects. For "orange" (O), "white" (W), and "yellow" (Y), they are
OWY
OYW
YOW
YWO
WOY
WYO

 

[Dick]

So why would I need to multiply my answer by 3 then? I am confused about that.

The probability of the first ball yellow and the other two white is 1/22. The probability of the second ball yellow and the other two white is 1/22. The probability of the third ball yellow and the other two white is 1/22. What's the probability that any ball is yellow and the other two white?

 

[phinds]

The probability of the first ball yellow and the other two white is 1/22. The probability of the second ball yellow and the other two white is 1/22. The probability of the third ball yellow and the other two white is 1/22. What's the probability that any ball is yellow and the other two white?

Gee, Dick, why don't you just spoon-feed him the answer so that he doesn't have to learn anything? Sorry to get snippy, but I though the POINT of this formum for this kind of question was to encourage thought, not give answers.

 

[Dick]

Gee, Dick, why don't you just spoon-feed him the answer so that he doesn't have to learn anything? Sorry to get snippy, but I though the POINT of this formum for this kind of question was to encourage thought, not give answers.

That is a little snippy. lacrotix KNOWS what the answer is. I just spelled out explicitly what I said in post 2. I was actually looking at Halls' OWY post when I wrote it and thinking how little it helped. Not yours. Sorry.

 

[phinds]

That is a little snippy. lacrotix KNOWS what the answer is. I just spelled out explicitly what I said in post 2. I was actually looking at Halls' OWY post when I wrote it and thinking how little it helped. Not yours. Sorry.

Well, it's possible that I go overboard in the direction of only providing direction and clues and studiously trying to avoid actually providing infomation that the OP should be able to figure out if he/she is going to learn anything. It is perhaps inappropriate of me to feel insistent that others should use that same technique.

 

[Dick]

Well, it's possible that I go overboard in the direction of only providing direction and clues and studiously trying to avoid actually providing infomation that the OP should be able to figure out if he/she is going to learn anything. It is perhaps inappropriate of me to feel insistent that others should use that same technique.

It's a fine goal, keep it up. Sorry again if I stepped on your hint.

 

[lacrotix]

There are n! (n factorial) ways to order n objects. For "orange" (O), "white" (W), and "yellow" (Y), they are
OWY
OYW
YOW
YWO
WOY
WYO

Sorry to write too much and force you to rush-read my question HallsofIvy. So yeah... this doesn't help with the stated question. :D I'll condense my question next time so that it is easier for everyone to follow.

The probability of the first ball yellow and the other two white is 1/22. The probability of the second ball yellow and the other two white is 1/22. The probability of the third ball yellow and the other two white is 1/22. What's the probability that any ball is yellow and the other two white?

Now that is helpful! So I see: I multiply by three to my calculated probability of (1/22) since the yellow ball can be drawn as the first, second, and third ball. Thank you very much for the help Dick.

Gee, Dick, why don't you just spoon-feed him the answer ...

Again, sorry phinds for not being clear in my original post. I didn't make it too clear that the answer the textbook gave was (3/22), that I understood the approach to the question, but didn't understand how to get to the middle. But thanks to Dick, now I do. :)


Thanks to everybody who has or has tried to help me. You guys are awesome!