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…
Best Answer
The number of cominations is not $_{16}C_2$ it must be $_{16}C_7$
Right?