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There are three bags each containing 5 red and 5 black balls. One ball is transferred blindly from the first to the second bag, and then one ball is transferred blindly from the second to the third. A ball is now drawn from the third bag. Find the probability that the drawn ball is red.

No idea how to proceed. Please help.

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    Case by case as in ball transferred from b1 to b2 is blue and b2 to b3 is red ball transferred from b1 to b2 is blue and b2 to b3 is blue etc – 2017-02-06

2 Answers 2

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Quick and simple: 50%

Since there's no difference in the amount of black and red balls the two end results (6 reds 5 blacks and 5 reds 6 blacks) are just as likely.

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$\Pr(red) $

$= \Pr(red|$transferred ball to third bin is red$)\Pr($transferred ball to third bin is red$) + \Pr(red|$transferred ball to third bin is black$)\Pr($transferred ball to third bin is black$) $

$= \frac{6}{11}\Pr($transferred ball to third bin is red$) + \frac{5}{11}\Pr($transferred ball to third bin is black$)$

$=\frac{6}{11}\Big(\Pr($transferred ball to third bin is red$|$transferred ball to second bin is red$)\Pr($transferred ball to second bin is red$) + \Pr($transferred ball to third bin is red$|$transferred ball to second bin is black$)\Pr($transferred ball to second bin is black$)\Big) + \frac{5}{11}\Big(\Pr($transferred ball to third bin is black$|$transferred ball to second bin is red$)\Pr($transferred ball to second bin is red$) + \Pr($transferred ball to third bin is black$|$transferred ball to second bin is black$)\Pr($transferred ball to second bin is black$)\Big)$

$=\frac{6}{11}\big(\frac{6}{11}\frac{1}{2} + \frac{5}{11}\frac{1}{2} \big) + \frac{5}{11}\big(\frac{5}{11}\frac{1}{2} + \frac{6}{11}\frac{1}{2} \big) = \frac{1}{2}$