# Thread: Probability Question-Not too difficult

1. ## Probability Question-Not too difficult

The question is:
John needs to choose randomly 20 chocolate coins from a jar containing 65 coins.
30 of the coins in the jar are green, 15 are yellow and 20 are red.

The parts I can't understand are:

Case 2 is: If all the coins from the same color are identical, how many different chioces are there ?

Part B: What is the probability that between the 20 coins that john will choose there will be 11 green coins and 4 yellow coins excatly?

My try:

About Case 2- It's the number of soloutions to the equation:
$\displaystyle x_{1} + x_{2} +x{3} = 20$ when 0<=x1<=30,
0<=x2<=15, 0<= x3<= 20... But how can I calculate this number?

About part B: I think the solution is $\displaystyle \frac { \frac{11}{30} \frac{4}{15} \frac{5}{20} }{ (65 over 20) }$ but I'm not so sure...I'll be delighted to get some verification on this ....

2. Originally Posted by WannaBe
The question is:
John needs to choose randomly 20 chocolate coins from a jar containing 65 coins.
30 of the coins in the jar are green, 15 are yellow and 20 are red.

The parts I can't understand are:

Case 2 is: If all the coins from the same color are identical, how many different chioces are there ?

Part B: What is the probability that between the 20 coins that john will choose there will be 11 green coins and 4 yellow coins excatly?

My try:

About Case 2- It's the number of soloutions to the equation:
$\displaystyle x_{1} + x_{2} +x{3} = 20$ when 0<=x1<=30,
0<=x2<=15, 0<= x3<= 20... But how can I calculate this number?

About part B: I think the solution is $\displaystyle \frac { \frac{11}{30} \frac{4}{15} \frac{5}{20} }{ (65 over 20) }$ but I'm not so sure...I'll be delighted to get some verification on this ....

Hi WannaBe,

this is a question dealing with ARRANGEMENTS (also known as "permutations") and SELECTIONS (also known as "combinations").

Applying logic to the first question,
the 20 chocolate coins can be either red or green.
They cannot be yellow as the coins could not then all be the same colour.

Since there are only 20 red coins, we'd have to select all 20 of them to have all red coins.
Given that they are identical (in other words, we could distinguish between them if they were numbered or different sizes etc), we will not notice any difference between various arrangements of them.
Hence the 20 reds is 1 choice of 20 or 1 selection of 20.

It's a different story with the greens.
There are 30 of these.

On your calculator, you have buttons to calculate $\displaystyle Np_R$ and $\displaystyle Nc_R$

The $\displaystyle Nc_R$ button calculates numbers of "selections",
while the $\displaystyle Np_R$ button calculates numbers of arrangements.

You can also make these calculations yourself using "factorials".

R is a subset of N.

$\displaystyle Np_R=\frac{N!}{(N-R)!}$

$\displaystyle Nc_R=\frac{N!}{(N-R)!R!}$

Hence the number of ways of choosing 20 of the green coins from 30 is $\displaystyle 30c_{20}$ which may also be written $\displaystyle \binom{30}{20}$

So, you must calculate that.

Part B

What is the probability of choosing 11 greens, 4 yellows and 5 reds ?

First, in how many ways can we choose 11 greens from 20
(imagine they are all assigned different numbers so we can distinguish them).
This is $\displaystyle 30c_{11}$

There are $\displaystyle 15c_4$ ways to choose 4 yellows

and there are $\displaystyle 20c_5$ ways to choose 5 reds.

In total there are $\displaystyle 65c_{20}$ ways to choose 20 coins from 65.

Any group of greens can go with any group of yellows.
Hence we multiply together the number of groups of both to find the number of selections of 11 green and 4 yellows.

All of these larger groups can go with any of the red groups.

Hence we multiply again.

So the probability is $\displaystyle \frac{number\ of\ 20G4Y5R\ groups}{65c_{20}}$

The second part is completely understandable and your answer is what I meant to write in my first msg but I didn't know how to make the binom sign in Latex...

I have to disagree with you about the first part:
Although there are only 15 yellow coins, how come we can't choose:
5 yellow + 6 red+ 9 green coins? We sure can! Actually it is my difficulty in this question...I can't figure out how to calculate the number of soloutions for the equation I wrote in my first msg... The soloution for the equation will be excatly as the number of selections in the first part...

Hope you'll be able to continue guiding me...

Thanks a lot again!

4. Originally Posted by WannaBe
The second part is completely understandable and your answer is what I meant to write in my first msg but I didn't know how to make the binom sign in Latex...

I have to disagree with you about the first part:
Although there are only 15 yellow coins, how come we can't choose:
5 yellow + 6 red+ 9 green coins? We sure can! Actually it is my difficulty in this question...I can't figure out how to calculate the number of soloutions for the equation I wrote in my first msg... The soloution for the equation will be excatly as the number of selections in the first part...

Hope you'll be able to continue guiding me...

Thanks a lot again!
Hi WannaBe,

Yes, i didn't read the first question right,
it's not asking for "if all 20 coins are the same colour".

Binomial in Latex is \binom{65}{20} for $\displaystyle \binom{65}{20}$

5. So can you guide me for the first one? How should I solve it?

Thanks

6. The alternatives could be listed

all green
20g....1

all red
20r....1

green and red together only
19g, 1r
18g, 2r

down to 1g, 19r That's a list of 19

green and yellow together only

19g, 1y
18g, 2y

down to 5g, 15y That's a list of 15

red and yellow .... same, a list of 15

green, yellow and red together

18g, 1r, 1y

17g, 1r, 2y
17g, 2r, 1y

16g, 1r, 3y
16g, 2r, 2y
16g, 3r, 1y

increasing by 1 each time until we get to

4g, 1r, 15y
4g, 2r, 14y.... a list of 15

3g, 2r, 15y
3g, 3r, 14y.....a list of 15

2g, 3r, 15y
2g, 4r, 14y....a list of 15

g, 4r, 15y
g, 5r, 14y......a list of 15

The total is 1+1+19+15+15+(1+2+3+....+14)+15(4)

$\displaystyle =21+\frac{14(15)}{2}+6(15)=21+7(15)+6(15)=21+13(15 )$

There is probably another way of doing it, though!

7. Originally Posted by WannaBe
So can you guide me for the first one? How should I solve it?
Here is one way.
If we expand the polynomial $\displaystyle \left( {\sum\limits_{k = 0}^{20} {x^k } } \right)^2 \left( {\sum\limits_{k = 0}^{15} {x^k } } \right)$ the coefficient of $\displaystyle x^{20}$ is $\displaystyle 210$.
That is the number of ways to select the twenty coins.

8. Wow thanks to both of you!