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Balls are drawn repeatedly and with replacement from a bag containing $60$ white and $30$ black balls. What is the probability of drawing the third white ball before the second black ball?

Can somebody help me with this?

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  • $\begingroup$ In order for this to happen, you must already have drawn two white balls and either 0 or 1 black ball. $\endgroup$ – user247327 Sep 10 '17 at 12:50
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The only possible scenarios that allow one to draw the third white ball before the second black ball are

⚪⚪⚪

⚫⚪⚪⚪

⚪⚫⚪⚪

⚪⚪⚫⚪

Compute the probability for each of these and add them to get your answer.

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