help asap please

There are 6 green balls and 4 red balls, making 10 total balls. The probability of picking a green ball first is \(\frac{6}{10}\) .

Since there are replacements, there will be still 10 balls in the bag. Now, for the second draw, you pick green again; making the probability, \(\frac{5}{10}\) . So, the probability of picking a green ball both times is \(\frac{6}{10}*\frac{5}{10}=\frac{30}{100}=\frac{3}{10}\) .

Next, the probability of picking a red ball is \(\frac{4}{10}\) . Doing the same as before, we get: \(\frac{4}{10}*\frac{3}{10}=\frac{3}{25}\) .

I'm leaving the next step up to you, to solve... What can we do after this?

Why is the probability of a second ball 5/10 when its color is just recorded and put back in the bag? You still have 6 green balls and 4 red balls. So the second draw should still be 6/10. No?.

OK, good man. Now, he says "What is the probability the two balls are the same color?" So, I think you have to add up the two probabilities together to answer his question: 9/25 + 4/25 =13/25 = 52%.