Probability, nearly solved

Question: Two distinct even numbers are selected at random from the first ten even numbers greater than zero. What is the probability that the sum is 30?

For this one, I have the denominator as 10! because there are 10 numbers. Then the numerator is what I'm not certain about. I know there are 4 ways to get a sum of 30 with those 10 numbers: 10+20, 8+22, 18+12, and 14+16. Since those numbers can be chosen in 2 different ways I did 8! for the numerator. Is that correct? When I solved 8!/10! I got 0.011. Have I gone about solving this correctly?

I also, have one more question that I have solved but not sure if I've done so correctly.

Question: A batch consits of 12 defective coils and 88 good ones. Find the probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made.

For the numerator I did 76 choose 2, and got 2850. I did this because this is the number of working coils there are. Then for the denominator I did 88 chose 2, and got 3828. I did this because this is the total number of coils. I divided and got 0.75. Have I gone about solving this correctly?

Thank You!

Hello, chelsea88!

Sorry, your approach is wrong on both problems.

Two distinct even numbers are selected at random
from the first ten even numbers greater than zero.
What is the probability that the sum is 30?

For this one, I have the denominator as 10! because there are 10 numbers.
No, that's if you want all 10 numbers in a particular order.

Then the numerator is what I'm not certain about.
I know there are 4 ways to get a sum of 30 with those 10 numbers: 10+20, 8+22, 18+12, and 14+16. . Right!
Since those numbers can be chosen in 2 different ways I did 8! for the numerator. . No

Click to expand...

\(\displaystyle \text{There are: }\:\text{10 choose 2} \:=\:_{10}C_2 \:=\:{10\choose2} \;=\;\frac{10!}{2!\,8!}\:=\:45\text{ possible pairs of even numbers.}\)

\(\displaystyle \text{There are }4\text{ pairs with a sum of 20.}\)

\(\displaystyle \text{Therefore: }\(\text{sum of 20}) \;=\;\frac{4}{45}\)

A batch consits of 12 defective coils and 88 good ones.
Two coils are selected with replacement.
Find the probability of getting two good coils.

For the numerator I did 76 choose 2, and got 2850. . Why 76 ?
I did this because this is the number of working coils there are. . no

Then for the denominator I did 88 chose 2, and got 3828. . no
I did this because this is the total number of coils.

Click to expand...

Since the sampling is done with replacement, the probabilites remain the same at each selection.

\(\displaystyle \text{There are 100 coils: }\;\begin{Bmatrix} \text{88 good} \\ \text{12 def.} \end{Bmatrix}\)

\(\displaystyle \text{Hence: }\(\text{good}) \:=\:\frac{88}{100} \:=\:\frac{22}{25}\)


\(\displaystyle \text{Therefore: }\(\text{two good}) \;=\;\frac{22}{25}\cdot\frac{22}{25} \;=\;\frac{484}{625}\)

darn it! but thank you. the reason i did 76 was because I had a similar problem where I had to find the numbers of ways to choose a selection with no defective products, and to do that i had to subtract the defective amount from the original amount, then do nCr. so i subtracted 12 from 88 to get 76, thinking that was the number of ways to get a non defective product.
Thank you