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- Oct 2nd 2011, 04:29 AM #1

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## About pdf

Hello. There are 3 boys, 3 pictures and 3 car names.

A task is just to connect car names into their pictures.

If a boy connects car names into pictures randomly, what is a probability function

on number Y of correct choises?

Let's see, 3 pictures, 3 cars and because it is enough that one boy connects, so

that would be: {1,2,3,4,5,6,7} and sum of that is 28.

( Unless, we have to deal correct choises: 3 pictures = 1/3. 3 car names = 1/3. ...)

Next step is division.

Thus, 1/28, 2/28, 3/28, 4/28, 5/28, 6/28, 7/28.

2/28 = 1/14, 4/28 = 1/7 and 7/28 = 1/4.

Hence, pdf ( probability density function ) on number Y of correct choises is

{1/28, 1/14, 3/28, 1/7, 5/28, 6/28, 1/4}

- Oct 2nd 2011, 05:31 AM #2
## Re: About pdf

- Oct 2nd 2011, 06:25 AM #3

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## Re: About pdf

I am not sure. But maybe you understood correctly my question.

I'm lookin for pdf ( I mean probability density function ).

And yes, any boy can connect car names into pictures randomly.

And any boy can get all three correct.

That {0,1,2,3,4,5,6,7,8,9}; will that be 3 boys + 3 cars + 3 pictures = 9

But is that {0,1,2,3,4,5,6,7,8,9} a probability density function on correct choises?

Because I know that pdf cannot be just a number.

- Oct 2nd 2011, 06:56 AM #4

- Oct 2nd 2011, 08:17 AM #5

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- Oct 2nd 2011, 09:19 AM #6

- Oct 2nd 2011, 11:29 AM #7

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- Oct 3rd 2011, 04:01 AM #8

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## Re: About pdf

P(Y = 1) =

0 0 1 1^2

1 - - - = ----

6 6 6 6^3

P(Y = 2) =

1 0 1 1^2

1 - - - = ---

6 6 6 6^3

P(Y = 3) =

1 1 1 1^3

1 - - - = ---

6 6 6 6^3

P(Y = 4) =

3 1 0 3^1

2 - - - = ---

6 6 6 6^3

P(Y = 5) =

3 1 1 3^1

2 - - - = ---

6 6 6 6^3

P(Y = 7) =

3 3 1 3^2

3 - - - = ---

6 6 6 6^3

This is really hard, but I truly tried to think this over. I'm afraid this is still wrong?

- Oct 3rd 2011, 04:20 AM #9

- Oct 3rd 2011, 10:56 AM #10

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- Oct 3rd 2011, 11:08 AM #11

- Oct 4th 2011, 07:01 AM #12

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## Re: About pdf

I can't understand more than this. I hope this is correct stuff;

Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3

Y = 4 in three ways: 103, 130, 301, 310, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3

Y = 5 in three ways: 113,131,311 P(Y = 5) = 3 * 3 * 3 / 6^3

Y = 6 in three ways: 330, 033, 303 P(Y = 6) = 3 * 3 * 3^2 / 6^3

Y = 7 in three ways: 331, 133, 313. P(Y = 7) = 3 * 3 * 3^2 / 6^3

So is that function

P = {111, 003, 300,103, 130, 301, 310, 031, 013, 113, 131,311,330, 033, 303, 331, 133, 313} ?

- Oct 5th 2011, 07:57 AM #13

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## Re: About pdf

I still don't know what is a probability density function on correct choises

IS IT THESE;

Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3

Y = 4 in three ways: 103, 130, 301, 310, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3

Y = 5 in three ways: 113,131,311 P(Y = 5) = 3 * 3 * 3 / 6^3

Y = 6 in three ways: 330, 033, 303 P(Y = 6) = 3 * 3 * 3^2 / 6^3

Y = 7 in three ways: 331, 133, 313. P(Y = 7) = 3 * 3 * 3^2 / 6^3

OR THIS:

P = {111, 003, 300,103, 130, 301, 310, 031, 013, 113, 131,311,330, 033, 303, 331, 133, 313} ?

OR is it like this;

Y = 3 = {111, 003, 300}

Y = 4 = {103, 130, 301, 310, 031, 013} and so on. ??