Cap Probability Problem with only 2 colored balls

An baseball cap has 2 red balls and 8 blue balls. You take a ball, record the color, then put it back in the cap. You take a second ball, record the color and then put it back as well.

What is the probability of selecting two red balls?
What is the probability of selecting two balls of the same color?
What is the probability of selecting two balls of different colors?
What is the probability of selecting a red ball on the second pick?
What is the probability of selecting no more than one blue ball?

my work:

2 red balls - (2/10) * (2/10) = 4/100
2 balls of the same color - (2/10) * (2/10) + (8/10)*(8/10) = 68/100 ? i wasn't sure about this one.
2 balls of different colors - (8/10) *(2/10) + (2/10) *(8/10) = 32/100
red ball on 2nd pick - (8/10)*(2/10) + (2/10)*(2/10) = 20/100
no more than one blue ball - i wasn't sure at all.

I'm not really sure what the formulas are for the last four questions. If i knew the formulas I could do it myself. I tried reading the other problems listed here, but they didn't go over only 2 colors, or the same questions i had. any help please? thanks!

cellobabee said:

An baseball cap has 2 red balls and 8 blue balls. You take a ball, record the color, then put it back in the cap. You take a second ball, record the color and then put it back as well.

What is the probability of selecting two red balls?
What is the probability of selecting two balls of the same color?
What is the probability of selecting two balls of different colors?
What is the probability of selecting a red ball on the second pick?
What is the probability of selecting no more than one blue ball?

my work:

2 red balls - (2/10) * (2/10) = 4/100(CORRECT)
2 balls of the same color - (2/10) * (2/10) + (8/10)*(8/10) = 68/100 ?(CORRECT)
2 balls of different colors - (8/10) *(2/10) + (2/10) *(8/10) = 32/100(CORRECT)
red ball on 2nd pick - (8/10)*(2/10) + (2/10)*(2/10) = 20/100(NOT CORRECT)

Click to expand...

The probability of a red ball on the second draw is the same as on the first draw.

What is the probability of selecting no more than one blue ball?
\(\displaystyle 1 - \frac{64}{100}\) the opposite of two blue balls.

pka said:

The probability of a red ball on the second draw is the same as on the first draw.

Click to expand...

if that is the case, the probability of a red ball on the 1st draw is 2 out of 10. isn't that the same as 20 out of 100?

or do i need to multiply 2/10 and 2/10 again to get 40/100?

thanks!

cellobabee said:

pka said:

The probability of a red ball on the second draw is the same as on the first draw.

Click to expand...

if that is the case, the probability of a red ball on the 1st draw is 2 out of 10. isn't that the same as 20 out of 100?

Click to expand...

Ha! I guess it is. I guess I wonder why not reduce it?

Yes, I messed up. I forgot to multiply by 2. DUH. I can't believe I done that.

Prob. of no more than 1 blue:

\(\displaystyle 2(8/10)(2/10)+(2/10)^{2}=\frac{9}{25}\)

Another way of counting it.