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# Thread: Math question, pissing me off.....

1. ## Math question, pissing me off.....

There are 3 oranges 5 apples and 2 bananas in a bag, what is the probablility of choosing one apple, one banana, and one orange. I swear to god the books answer key is wrong and I am pretty pissed off right now, what do you guys get?

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50% chance of getting an apple, 30% chance of getting an orange, 20% chance of getting a banana.... seems simple enough lol...

Or, do you mean what is the chance of picking 1 of each at the same time?
Your question lacks some detail here... fill me in...
Last edited by Blown_SC; 10-01-2004 at 01:16 AM.

3. you are choosing 3 items at once, whats the probability that you will get exactly one apple one orange and one banana, with 7 items remaining.
I know its not that simple SC, gimme a little credit

Bas ass avatar by the way, Im going out to wash mine right now

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Originally Posted by JDawg1536
you are choosing 3 items at once, whats the probability that you will get exactly one apple one orange and one banana, with 7 items remaining.
I know its not that simple SC, gimme a little credit

Bas ass avatar by the way, Im going out to wash mine right now
Bro, do you have to pick 3 items each time you enter the bag?
Is it possible to pick only one item? Or do you have to pick 3 items no matter what...

5. trinomial expansion maybe? thats what it woudl apear like

you know... (a+b+c)^n a=(3/10) b=(5/9) c=(2/8)

you would want the one coresponding to X(abc)

which would be actually 6*a*b*c

so it should be 180/720 = ~ 1/4

Its a non-replaced event with non-ordering

Theres a simpler forumula...

I dont remeber it..its something like

x!y!z! / a!n! a!b! or something like that.., basically a computation or permutation i hate statistics

6. Originally Posted by Billy_Bathgate
trinomial expansion maybe? thats what it woudl apear like

you know... (a+b+c)^n a=(3/10) b=(5/9) c=(2/8)

you would want the one coresponding to X(abc)

which would be actually 6*a*b*c

so it should be 180/720 = ~ 1/4

Its a non-replaced event with non-ordering

Theres a simpler forumula...

I dont remeber it..its something like

x!y!z! / a!n! a!b! or something like that.., basically a computation or permutation i hate statistics
ok i get using the trinomial expansion, but where are you getting x and n from?

Also SC, it wouldnt matter how you pick them.

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[(5C1) (3C1) (2C1)] / (10C3) = 0.25
Last edited by l6873; 10-01-2004 at 04:08 AM.

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Originally Posted by Billy_Bathgate
trinomial expansion maybe? thats what it woudl apear like

you know... (a+b+c)^n a=(3/10) b=(5/9) c=(2/8)

you would want the one coresponding to X(abc)

which would be actually 6*a*b*c

so it should be 180/720 = ~ 1/4

Its a non-replaced event with non-ordering

Theres a simpler forumula...

I dont remeber it..its something like

x!y!z! / a!n! a!b! or something like that.., basically a computation or permutation i hate statistics
I'm an idiot

9. 100% I would say... If i reach in a basket, I can 100% gurrantee everytime Im going to pull out 1 apple, 1 banana, and 1 orange!

max

10. Originally Posted by max2extreme
100% I would say... If i reach in a basket, I can 100% gurrantee everytime Im going to pull out 1 apple, 1 banana, and 1 orange!

max
I like strawberries.

11. The key here is without replacement so you "re-calculate" the probabilities after each so called pick because the pool shrinks. Kinda of like a game of cards, as you get cards dealt to you the probablies are changing. Like Bathgate, I can't remeber the **** formula and my old math book is all the way across the room so that's all I got..lol.

12. when the probability is like that, you multiply the chance of getting each individual one together.
Apple= .5
Orange= .3
Banana= .2
.5 X .3 X .2 = .03

There is a 3% chance that if you grab 3 items at the same time out of the bag, there will be one of each.

13. im glad i never had to take statistics, differential equations was annoying enough for me.

14. Originally Posted by Superhuman
when the probability is like that, you multiply the chance of getting each individual one together.
Apple= .5
Orange= .3
Banana= .2
.5 X .3 X .2 = .03

There is a 3% chance that if you grab 3 items at the same time out of the bag, there will be one of each.
No Go back to school!
j/p

15. Originally Posted by JDawg1536
There are 3 oranges 5 apples and 2 bananas in a bag, what is the probablility of choosing one apple, one banana, and one orange. I swear to god the books answer key is wrong and I am pretty pissed off right now, what do you guys get?
In total there is 10 fruits. If were not repalcaing them:

3/10 * 5/9 * 2/8 (one sequnce of picking out the fruit)

+

5/10 * 2/9 * 3/8 (another sequence)

+

2/10 * 3/9 * 5/8 (another)

+

3/10 * 2/9 * 5/8 (guess what?)

+

2/10 * 5/9 * 3/8 (yep you got it... another)

+

5/10 * 3/9 * 2/8 (and the final sequence)

= 1/24 + 1/24 + 1/24 + 1/24 + 1/24 + 1/24

= 3!(1/24) = 1/4 or 0.25 or 25% Whatever you like!

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Originally Posted by Tushe
In total there is 10 fruits. If were not repalcaing them:

3/10 * 5/9 * 2/8 (one sequnce of picking out the fruit)

+

5/10 * 2/9 * 3/8 (another sequence)

+

2/10 * 3/9 * 5/8 (another)

+

3/10 * 2/9 * 5/8 (guess what?)

+

2/10 * 5/9 * 3/8 (yep you got it... another)

+

5/10 * 3/9 * 2/8 (and the final sequence)

= 1/24 + 1/24 + 1/24 + 1/24 + 1/24 + 1/24

= 3!(1/24) = 1/4 or 0.25 or 25% Whatever you like!

http://forums.anabolicreview.com/sho...72&postcount=7

17. Originally Posted by l6873
[(5C1) (3C1) (2C1)] / (10C3) = 0.25
This is the correct way to go about answering this type of problem. Don't make statistics and probability overly complicated -- know the concepts, formulas and how to apply them. That way you don't take a whole page to do something that can be done in one line.

18. I'd say 25%.

19. 20 percent are you tarded or what..

20. Originally Posted by JDawg1536
ok i get using the trinomial expansion, but where are you getting x and n from?

Also SC, it wouldnt matter how you pick them.
X would be the coeficiant that comes down. In this case it is 6 it follows the inverted pyramid

the forumula i believe is (prior coefficient * degree of first variable power / cofficent place number)

n is the number of events, in this case 3

the other forumula, is basically the forumula for a computation or whatever it is called

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Originally Posted by Money Boss Hustla
I like strawberries.

hey Mr. Wizard there are no strawberries in the bag? geez

BB, can you make me feel any fukin dumber?

22. this thread sucks , so im not even gonna bother workin it out .

23. Didnt see, is that 3 separate pulls or 1 pull. if its 3 separate pulls, the answer is 1 in 4 chance of reaching in 3 separate times, and pulling out an apple, a banana, and an orance.

24. I dont know about anyone that has seen this thread but i hate math. im we-ta-ded

25. While this specific question can be answered intuitively, the way most people go about figuring out probabilities and statistics is dead wrong. That's why those of us 'in the know' refer to the lottery as a tax for the mathematically noninclined.

For example, take a look at the incidence of a rare disease. Say only 1 in 1000 adults is affliced with this rare disease for which a diagnostic test has been developed.

The test is such that when a person actually has the disease, a positive result will occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time.

If a random selected individual is tested and the result is positive, what is the probability that the individual has the disease?

Think about that for a minute. The test seems pretty accurate. Everyone knows that 99% is accurate, right?

The probability that the individual has the disease is 4.7%.

Lots of probs and stats are very counterintuitive... be careful when solving them.