Statistics

[K.I.L.E.R]

If you have a 10 multiple choice question test, with each question having 4 options each. How many different answer combinations can you get?

Now, statistically speaking, what are the chances of 2 random tests having 8 questions answered identically?

Show workings.

a:

nCr = 40!/36!4! = 91390 combos
10 questions, 4 options per question, 10*4 = 40

b:
nPr = nCr * p2 * (1-p)^n-2
nCr = 10c8
n = 10 questions, r = 8 done in a certain order.
p = 1/4 (one option out of 4)

1 test getting a certain combo of 8 questions:
0.006257

For another test getting same combo:
0.006257 ^ 2?

Right or wrong?

 

[Simon F]

a:

nCr = 40!/36!4! = 91390 combos

No. wrong.
I won't give you the answer but imagine there were only two choices per question and that those were "bits". How many different values would you have?

 

[AlNom]

40c10

 

[K.I.L.E.R]

Whoops I noticed my retarded error.
10c4.

210 combos.

No. wrong.
I won't give you the answer but imagine there were only two choices per question and that those were "bits". How many different values would you have?

 

[Simon F]

Whoops I noticed my retarded error.
10c4.

NO!

You're being mislead by the word "combinations". Ok, here's a big hint. You have a 5 bit value. How many different values of bits can you have?

 

[K.I.L.E.R]

10p4

It has to be because that takes into account probability of a option being chosen.

p = 1/4.

NO!

You're being mislead by the word "combinations". Ok, here's a big hint. You have a 5 bit value. How many different values of bits can you have?

 

[Tim Murray]

fine, I take solace in the fact that I'm right.

ps. Simon's right, stop focusing on the word combination. combination does not imply the use of n choose k.

 

[Simon F]

....

So he doesn't get to do his own homework?

 

[Jabbah]

So he doesn't get to do his own homework?

Lets see if he can work out why 4^10 now...

 

[K.I.L.E.R]

I already seen the answer.
4^10.

4^10 doesn't really make sense though.

 

[K.I.L.E.R]

We only studied statistics under text book conditions.
If the question isn't stated in the form of a textbook question it really raises issues.

fine, I take solace in the fact that I'm right.

ps. Simon's right, stop focusing on the word combination. combination does not imply the use of n choose k.

 

[Jabbah]

How many ways of answering one 4 option question?
How many ways of answering two 4 option questions?

 

[epicstruggle]

I already seen the answer.
4^10.

4^10 doesn't really make sense though.

why not work your way from a smaller problem to what your doing?

Start with 3 questions, 2 possible answers. increase the number of question and/or # of answers to see what happens to the result.

epic

 

[Tim Murray]

I already seen the answer.
4^10.

4^10 doesn't really make sense though.

if you didn't understand this, stop posting here and go to a TA or the professor. get actual help.

 

[K.I.L.E.R]

1/4 for q1
(1/4)^10

However that implies you can only select a single option.
So:

1/4 * 4 ^ 10

How many ways of answering one 4 option question?
How many ways of answering two 4 option questions?

 

[K.I.L.E.R]

Or you can scoff up a result.
Gimme a break, it's 1:23am down here.

if you didn't understand this, stop posting here and go to a TA or the professor. get actual help.

 

[Jabbah]

1/4 for q1

Thats the probability, not the number of different possible answers.

 

[K.I.L.E.R]

There are 4 possible answers.
That's why I multiplied 1/4 by 4.

Thats the probability, not the number of different possible answers.

 

[Tim Murray]

ten independent events. four options per event. 4^10.

seriously, get off the internet, read the textbook about seven more times straight through, and then go talk to the professor. you are completely screwed if you don't. this is not hard. it's incredibly fundamental. b is slightly harder but it's still not bad, but if you have trouble with A, you desperately need help.

 

[Jabbah]

4 different ways to answer q1
4 different ways to answer q2

how many different ways to answer q1&q2?