# Math Help - 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}

Originally Posted by TractorGreen
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 choices?
Surely, the outcomes are $\{0,1,2,3,4,5,6,7,8,9\}$.
It could be that all three could be correct: $\mathcal{P}(Y=9)=\frac{1}{6^3}$.
Because any boy can get all three correct with probability $\frac{1}{6}$.
Have I misunderstood the question?

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.

Originally Posted by TractorGreen
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.
Each boy can get $\{0,1,3\}$ note it is impossible to get a $2$.
$\mathcal{P}(0)=\frac{2}{6},~\mathcal{P}(1)=\frac{3 }{6},~\mathcal{P}(3)=\frac{1}{6}$.

Because no one can get a 2, it is impossible to get 8.

We can get a 6 by $033,~ 303,~330$.

So you have to figure out all possible sums.

Okay. Here is my suggestion:

all possible sums for {0,1,3}

0 = 0, 00, 000
1 = 1, 01, 10, 001, 010
3 = 111, 003, 030, 003

Thus, a probability density function on correct choises is

{0,00,000,1,01,10,001,010,111,003,030,003}

Originally Posted by TractorGreen
Okay. Here is my suggestion:
all possible sums for {0,1,3}
0 = 0, 00, 000
1 = 1, 01, 10, 001, 010
3 = 111, 003, 030, 003
Thus, a probability density function on correct choises is
{0,00,000,1,01,10,001,010,111,003,030,003}
I have no idea what any of that means.
But $\mathcal{P}(Y=6)=3\frac{2}{6}\frac{1}{6}\frac{1}{6 }$.
$\mathcal{P}(Y=0)=\frac{2^3}{6^3}$ because that only happens if $000$.

Now you have to work out how $Y=1,2,3,4,5,\&~7$.

Hello.

This is difficult, but I sure try.

0 1 1 1^3
P(Y = 1) = 1 - - - = ----
6 6 6 6^3

1 0 1 1^3
P(Y = 2) = 1 - - - = ---
6 6 6 6^3

1 1 1 1^3
P(Y = 3) = 1 - - - = ---
6 6 6 6^3

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?

Each boy can get $\{0,1,3\}$ note it is impossible to get a $2$.
$\mathcal{P}(0)=\frac{2}{6},~\mathcal{P}(1)=\frac{3 }{6},~\mathcal{P}(3)=\frac{1}{6}$.
Note $Y=1$ in three ways: $100,~010,~001$ so $\mathcal{P}(Y=1)= 3\cdot\frac{3\cdot2^2}{6^3}$.
Note $Y=2$ in three ways: $110,~011,~101$ so $\mathcal{P}(Y=2)= 3\cdot\frac{3^2\cdot2}{6^3}$.

Now you must build each of the remaining cases.

This is 3 and 4. I hope these are right.

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

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

Originally Posted by TractorGreen
This is 3 and 4. I hope these are right.
Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3
Y = 4 in three ways: 301, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3
You missed one for Y=3: $030$.

You missed three for Y=4: $103,~130,~310$.

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} ?

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. ??