Question

An urn contains 6 red balls and 3 blue balls. One ball is selected at random and is replaces by a ball of the= other color. A Second ball is then chosen. What is the conditional probability that the first all selected is red, given that the second ball was red?

Answer 1

R_1 - First ball Red, R_2-second ball is red
P(R_1|R_2)=P(R_1 AND R_2)/P(R_2)
P(R_1 AND R_2) = 6/9 * 5/9 = 30/81=10/27
P(R_2) = 6/9 * 5/9 + 3/9 * 6/9 = 16/27 P(R_1|R_2)=(10/27)/(16/27) = 5/8.
I got it wrong. Why?

Answer 2

well you need to use the red ball as as your second

Answer 3

if you know what i mean?

Answer 4

but there only 6 red balls you dont care about the the 3 balls

Answer 5

\[\large P(R _{1}\cap R _{2})=\frac{6}{9}\times\frac{5}{9}=\frac{30}{81}\]
\[\large P(B _{1}\cap R _{2})=\frac{3}{9}\times\frac{7}{9}=\frac{21}{81}\]
\[\large P(second\ is\ red)=\frac{30}{81}+\frac{21}{81}=\frac{51}{81}\]
\[\large P(first\ was\ red)=\frac{\frac{30}{81}}{\frac{51}{81}}=\frac{10}{17}\]

Answer 6

You still haven't explain why is it 7/9

Answer 7

If a blue ball is selected in the first draw, it is replaced by a red ball (the other color). This results in there being 2 blue balls and 7 red balls for the second draw. The probability of red in the second draw, in this case, is 7/9.

Answer 8

There only 6 red balls.

Answer 9

It should be obvious that the replacement balls are not part of the original 9 balls. So, if a blue ball is replaced by a red ball, that red ball is taken from a supply of extra balls. That is the reason for it being possible to have 7 red balls.

Answer 10

I don't get it still

Answer 11

What part don't you get?

Answer 12

7/9

Answer 13

Do you accept that there must be additional balls outside the urn to make the exchange of color possible? There is always a total of 9 balls inside the urn, however the ratio of red to blue balls changes after the first draw, when the exchange of color is made.

Answer 14

If the first ball that is drawn is blue, then the blue ball is kept outside the urn and a red ball from the quantity initially outside the urn is put into the urn. This results in there being only 2 blue balls and 7 red balls inside the urn. Therefore when the second draw is made the probability of drawing a red ball is given by:
\[\large P(red)=\frac{number\ of\ red\ balls}{total\ number\ of\ balls}=\frac{7}{9}\]

Answer 15

so the blue ball turns to a red ball?

Answer 16

Yes, if a blue ball is drawn first, the blue ball stays outside the urn and a red ball from a supply outside the urn is put into the urn.

Answer 17

That makes sense. I thought you said that the color just turned but you really just adding the color back. So this questions is reallu asking with replacement.

Answer 18

Yes, replacement with the other color.