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Six green balls and four red balls are in a bag. A ball is taken from the bag, its color recorded, then placed back in the bag. A second ball is taken and its color recorded. What is the probability the two balls are the same color?

Jul 2, 2018

#1
+4223
+2

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?

Jul 2, 2018
#2
+2

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?.

Jul 2, 2018
#3
+4223
+2

Yes, thank you, guest, forgot about that!!!

So, both are 6/10, thus $$\frac{6}{10}*\frac{6}{10}=\frac{9}{25}$$.

And for the red, $$\frac{4}{10}*\frac{4}{10}=\frac{4}{25}$$.

Thanks again, guest!

tertre  Jul 2, 2018
#4
+1

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%.

Jul 2, 2018
edited by Guest  Jul 2, 2018