Urn problem question. having trouble.

[Tangentjay]

The urn holds 300 blue & 700 green balls. Select 10 balls with replacement,


there are a total of 1k balls. This is a combination question I'm sure, but I am getting a big number when trying to solve this by Expressing the answers as fractions with exponents

Maybe my math is way off,

What is the probability all ten balls are blue? (300C10/ 1000C10) is this how I would do the math?

What is the probability at least one ball is green. Not sure how to do this one.

 

[Tangentjay]

The probability of selecting a blue ball =
300/1000 = 0.3.
The probability of selecting 10 blue balls with replacement = (0.3)^10 = 59049/10^10.
is what I got for the first answer

 

[HallsofIvy]

Since your question specifically says "without replacement" there is no reason to even mention "with replacement"! Initially there are 1000 balls, 300 green and 700 blue. The probability the first ball chosen is blue is 700/1000. There are then 999 balls left, 699 of them blue. The probability the second ball chosen is also blue is 699/999. There are then 998 balls left, 698 of them blue. The probability the third ball chosen is also blue is 698/998. Continuing like that, it should be obvious that the numerator and denominator of the fraction decrease by one each time. The probability that all 10 balls will be blue is \(\displaystyle \frac{700}{1000}\frac{699}{999}\frac{698}{998}\cdot\cdot\cdot\frac{691}{991}\).

You can, if you like, simplify the notation. The numerators are (700)(699)(698)...(691) and that product can be written \(\displaystyle \frac{700!}{690!}\). Similarly, the denominators aree (1000)(999)(998)...(991) and that can be written \(\displaystyle \frac{1000!}{990!}\).

The answer can be written \(\displaystyle \frac{990!700!}{1000!690!}\).

 

[Dr.Peterson]

Since your question specifically says "without replacement"

I'm confused. Where does it say "without replacement"?

The urn holds 300 blue & 700 green balls. Select 10 balls with replacement,

there are a total of 1k balls. This is a combination question I'm sure, but I am getting a big number when trying to solve this by Expressing the answers as fractions with exponents

Maybe my math is way off,

What is the probability all ten balls are blue? (300C10/ 1000C10) is this how I would do the math?

What is the probability at least one ball is green. Not sure how to do this one.

Since it says "with replacement", this is not a combination problem (unless you omitted another part that is without replacement). But your first answer would be correct in the latter case. I won't say anything about the second part without being sure of the problem.

The probability of selecting a blue ball =
300/1000 = 0.3.
The probability of selecting 10 blue balls with replacement = (0.3)^10 = 59049/10^10.
is what I got for the first answer

This is correct; 0.3^10 = 0.0000059049.

What is the probability at least one ball is green. Not sure how to do this one.

Now, assuming it really does say with replacement, look for the probability that no balls are green. What does it mean if none are green?

If it is not true that none are green, then at least one must be green ...