An urn contains 3 red balls and 7 green balls A ball is to b

An urn contains 3 red balls and 7 green balls. A ball is to be drawn at random if it is red it is replaced. if it is green we put it back into the urn and draw again. Find the probablity that both balls drawn are the same color?

How would I solve this problem?

Solution

The phrasing is odd that both are replaced (put back after drawing), but we do it anyway.

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P(same color) = P(red 1st)P(red 2nd) + P(green 1st)P(green 2nd)

As they are replaced, then the probability of red or green in the first and second are the same. Hence,

P(same color) = P(red 1st)P(red 2nd) + P(green 1st)P(green 2nd)

= (3/10)*(3/10) + (7/10)*(7/10)

= 0.58 [ANSWER]

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