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An urn contains one red and one black ball. Each time, a ball is drawn independently at random from the urn, and then returned to the urn along with another ball of the same colour. For example, if the first ball drawn is red, the urn will subsequently contain two red balls and one black ball.

What is the probability of observing the sequence r,b,b,r,r?

I believe the probability is as follows:
$1/2$ * $1/3$ * $1/2$ * $2/5$ * $1/2$ = $0.016665$

What is the probability of observing 3 red and 2 black balls?
$(5C2) = 10 * 0.016665 = 0.16665$

What is the probability of observing 7 red and 9 black balls?
$(1/2) * (2/3) * (3/4) * (4/5) * (5/6) * (6/7) * (7/8) * (1/9) * (2/10) * (3/11) * (4/12) * (5/13) * (6/14) * (7/15) * (8/16) * (9/17) = 0.00000514191$
$(16C2) = 120$
$0.00000514191 * 120 = 0.0006170292$

The last part is wrong, but I do not know why. I got the probability of a single sequence and multiplied it by the number of combinations. I was also wondering if there is a nicer way to solve it...

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  • $\begingroup$ I don't follow the second one. Why shouldn't the sequence be r,r,r,b,b? Why should that have the same probability as r,b,b,r,r? $\endgroup$ – saulspatz Jul 31 '19 at 12:19
  • $\begingroup$ The sequence can be any, it just has to contain 3 red balls and 2 black balls. So, P(r,r,r,b,b) + P(r,b,b,r,r) +... There are 10 such combinations. $\endgroup$ – Libor Zachoval Jul 31 '19 at 12:21
  • $\begingroup$ But why do you think all sequences have the same probability? $\endgroup$ – saulspatz Jul 31 '19 at 12:22
  • $\begingroup$ Well, a hint was provided that they do. Also, P(r,r,r,b,b) = P(r,b,b,r,r) = P(3 red balls, 2 black balls) $\endgroup$ – Libor Zachoval Jul 31 '19 at 12:26
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    $\begingroup$ @drhab The denominator is always the number of balls in the bag, so that doesn't depend on the sequence. The numerator is the number of the ball of the color drawn. (When you draw the first red ball it's $1$, when you draw the third black ball, it's $3$.) Thus, the numerator doesn't depend on the sequence, but only on the number of balls of each color. $\endgroup$ – saulspatz Jul 31 '19 at 12:38
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The number of cominations is not $_{16}C_2$ it must be $_{16}C_7$

Right?

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I made a mistake. It should be $(16C7)$ and so $16!/(7!*9!) = 11440$

$0.00000514191 * 11440 = 0.0588234504$

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  • $\begingroup$ I am sorry when I typed my answer your answer wasn't yet shown on my screen, maybe because of a connection error $\endgroup$ – Fareed Abi Farraj Jul 31 '19 at 12:27
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    $\begingroup$ Yes, it must have been, or maybe it takes a bit to upload the actual answer. But I do appreciate your help! $\endgroup$ – Libor Zachoval Jul 31 '19 at 12:29

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