Very dumb stats question

[Juice Box]

yes it is homework....but I think the computer grading may be incorrect....let me know

1. True/False: If a die is rolled three times, the chance of getting at least one ace is 1/6+1/6+1/6=1/2

that of coruse is false

The teacher then added the question "What is the chance of getting at least one ace on 3 rolls of a die?"

I can only assume an "ace" means rolling a one...in which case are the odds not 1/6???

I put that in and it says it is wrong. I also tried 1/2 just to see, but still no. what in the world is the problem here?

 

[Vegitto]

Do you have a TI calc? In that case, this is easy. A dice has 6 sides. Only one (1) of those sides can be up, and it has to be the 1. This process of rolling the dice is repeated 3 times. The question is; how many times would one have to roll this dice to be certain they have (at least) 3 ones?

Just put in this: (6 nCr 1)^3. That's the answer.

 

[Baked]

^ Cheater. They should ban all calculators in schools and make people figure out the formula and solution on paper.

 

[johnnqq]

i'm doing the same bullsh1t...MATH SUCKS!!

 

[Juice Box]

Originally posted by: Vegitto
Do you have a TI calc? In that case, this is easy. A dice has 6 sides. Only one (1) of those sides can be up, and it has to be the 1. This process of rolling the dice is repeated 3 times. The question is; how many times would one have to roll this dice to be certain they have (at least) 3 ones?

Just put in this: (6 nCr 1)^3. That's the answer.

did that.....gives me 216

1/216 is also incorrect

 

[dullard]

I find it odd to call "1" and "Ace". So by definition, I say the odds are zero.

But lets calculate the odds of not getting a "1".

Roll 1: Odds of not getting a "1" is 5/6.
Roll 2: Odds of not getting a "1" is 5/6.
Roll 3: Odds of not getting a "1" is 5/6.

Thus after 3 rolls, the odds of never getting a "1" is (5/6)*(5/6)*(5/6) = 57.9%. The odds of getting at least one "1" is 1 - 57.9% = 42.1%.

 

[Vegitto]

Oh, then it's still not so hard.
(6 nCr 1) is equal to ((6!)/(1!x((6-1)!)).
We'll call the outcome 'x'.
After we have ((6!)/(1!x((6-1)!)), just do x times x times x. (= (x(x(x))) )

 

[Vegitto]

Originally posted by: Juice Box

Originally posted by: Vegitto
Do you have a TI calc? In that case, this is easy. A dice has 6 sides. Only one (1) of those sides can be up, and it has to be the 1. This process of rolling the dice is repeated 3 times. The question is; how many times would one have to roll this dice to be certain they have (at least) 3 ones?

Just put in this: (6 nCr 1)^3. That's the answer.

did that.....gives me 216

1/216 is also incorrect

I had the same question (and answer) on a math test a week ago, only the dice had 8 sides to confuse us. I had the question right, this is the answer.

 

[Juice Box]

Originally posted by: dullard
I find it odd to call "1" and "Ace". So by definition, I say the odds are zero.

But lets calculate the odds of not getting a "1".

Roll 1: Odds of not getting a "1" is 5/6.
Roll 2: Odds of not getting a "1" is 5/6.
Roll 3: Odds of not getting a "1" is 5/6.

Thus after 3 rolls, the odds of never getting a "1" is (5/6)*(5/6)*(5/6) = 57.9%. The odds of getting at least one "1" is 1 - 57.9% = 42.1%.

winner! thank you good buddy, guess i was going about it the wrong way

42.1/100 works!

thanks!!

 

[dullard]

Originally posted by: Juice Box
winner! thank you good buddy, guess i was going about it the wrong way

thanks!!

You are welcome. In stats, you usually have to calculate the odds of something "not" happening. Then subtract from 1 to get the odds for it happening. It is a simple trick that many fall victim to.

 

[Vegitto]

Originally posted by: Juice Box

Originally posted by: dullard
I find it odd to call "1" and "Ace". So by definition, I say the odds are zero.

But lets calculate the odds of not getting a "1".

Roll 1: Odds of not getting a "1" is 5/6.
Roll 2: Odds of not getting a "1" is 5/6.
Roll 3: Odds of not getting a "1" is 5/6.

Thus after 3 rolls, the odds of never getting a "1" is (5/6)*(5/6)*(5/6) = 57.9%. The odds of getting at least one "1" is 1 - 57.9% = 42.1%.

winner! thank you good buddy, guess i was going about it the wrong way

42.1/100 works!

thanks!!

Huh? I don't get it. I posted the answer (which was assumed correct by me and my maths teacher), and it's wrong? How can this be? My world is collapsing! Oh noes!

Good for you .

 

[Minjin]

A way to think of it is to consider the possible ways you can roll a 1:

1 - not 1 - not 1
not 1 - 1 - not 1
not 1 - not 1 - 1

Find the probability of each and add them.

(1/6)(5/6)(5/6)
(5/6)(1/6)(5/6)
(5/6)(5/6)(1/6)

Add those three results and that should be the answer.

edit: nevermind, I read it as the probability of getting ONE 1

Mark

 

[dullard]

Originally posted by: Vegitto
Huh? I don't get it. I posted the answer (which was assumed correct by me and my maths teacher), and it's wrong? How can this be? My world is collapsing! Oh noes!

You posted an answer to your own question which has nothing to to with Juice Box's question. You are basically living in your own world here; a thread within a thread; an unknown inner thread that no one cares about.

 

[Vegitto]

Originally posted by: dullard

Originally posted by: Vegitto
Huh? I don't get it. I posted the answer (which was assumed correct by me and my maths teacher), and it's wrong? How can this be? My world is collapsing! Oh noes!

You posted an answer to your own question which has nothing to to with Juice Box's question. You are basically living in your own world here; a thread within a thread; an unknown inner thread that no one cares about.

Yay! I have made more enemies in a minute than I would normally do... In a minute ¬_¬...