chosing K elements from the pool of N with Z specific ones

[vclaire]

Hello all,
i'm not a student, just want to make my point to a friend in an argument but failed to calculate the probabilities.

We have a pool of 400 elements (let's say balls), where 4 elements are special (let's say red) and 396 elements are normal (let's say black). We are taking a random sample from this pool. The samplesize is 10.

I want to calculate 3 probabilities:
1. The probabitity that no red ball ends up in our sample.
2. The probability that exactly 1 red ball ends up in our sample.
3. The probability that all 4 red balls end up in our sample.

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The first was easy (i guess). In our first draw the probability of black ball is 396/400, in the second draw it is 395/399... in the 10. draw it is 387/391. If we multiply them, we get 90,3338%

	1. draw	2. draw	3. draw	4. draw	5. draw	6. draw	7. draw	8. draw	9. draw	10. draw
Number of black	396​	395​	394​	393​	392​	391​	390​	389​	388​	387​
Number of balls	400​	399​	398​	397​	396​	395​	394​	393​	392​	391​
Prob.	0,99​	0,989975​	0,9899497​	0,989924​	0,989899​	0,989873​	0,989848​	0,989822​	0,989796​	0,98977​
Prob. all:	0,903338319​

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The second one: i'm sure, there's an easier way, but i calculated the probability that i will draw a red ball at first 4/400, then only black balls 396/399, 395/398.... 388/391. Then multipied them, so at the end i got that the probability that i draw one red ball first and all the others are black is 0,933683%

	1. draw	2. draw	3. draw	4. draw	5. draw	6. draw	7. draw	8. draw	9. draw	10. draw
Number of red	4​
Number of black		396​	395​	394​	393​	392​	391​	390​	389​	388​
Number of balls	400​	399​	398​	397​	396​	395​	394​	393​	392​	391​
Prob.	0,01​	0,992481​	0,9924623​	0,992443​	0,992424​	0,992405​	0,992386​	0,992366​	0,992347​	0,992327​
Prob. all.	0,00933683​

Because the red ball can be anywhere in the 10 spots, i multiplied this probability by 10 and got that the probability of having exactly 1 red ball is 9,33683%
Please check if this calculation is correct, and i'm sure, there is an easier way, please, could you show me!

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The third one: I'm completly stuck! I tried to draw a tree like

but realized, that it would take a lot of time, to reach the 10. draw, because at the end it will have 2^10 rows.
Please, could you show me, how to calculate this!


Thank you for your help (and sorry for my english)

 

[pka]

I would calculate \(\displaystyle \dfrac{\dbinom{396}{10}}{\dbinom{400}{10}}\) Use this link to see results.

 

[Dr.Peterson]

Pka's answer is for question 1 (all black).

Your work for question 2 is good; look up the binomial distribution for confirmation.

For question 3, a method similar to pka's is good: the numerator will be the ways to choose the 4 red balls (of 4) and 6 black balls (of 396), and the denominator will be the ways to choose any 10 balls.

 

[vclaire]

Thank you for your answer pka and Dr.Peterson!

So basicly my three probabilities are:
First (all black): which is basicly (396 choose 10)/(400 choose 10)

Second (one red):

Third (four red): which is basicly

Did i get it right?

Thank you again for your help!

 

[Dr.Peterson]

Yes. Good work!