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Mathematics 16 Online
OpenStudy (anonymous):

There are five red balls and five blue balls in a bag. When a red ball is removed it is always replaced. When a blue ball is removed a coin is flipped. If heads appears the ball is not replaced. If tails appears, then 2 blue ones are placed into the bag. What is the probability that first three balls drawn are the same colour? I know the P(RRR) = .125 I cannot understand the process to get the blue.

OpenStudy (anonymous):

we can compute this

OpenStudy (anonymous):

you want probability that all three are blue first one is blue with probability \(\frac{1}{2}\) second one we break in two parts, if the coin is heads, probability second one is heads is \(\frac{4}{10}\) if it is tail, probability second one is blue is \(\frac{6}{10}\) so we multiply and add \[\frac{1}{2}\times \frac{4}{10}+\frac{1}{2}\times \frac{6}{10}\] to get the probability second is blue

OpenStudy (anonymous):

hmm a quick calculation tells me probability second one is blue is also \(\frac{1}{2}\)

OpenStudy (anonymous):

im doubting the P(RRR) now. |:

OpenStudy (anonymous):

no that is right \[(\frac{1}{2})^3=\frac{1}{8}\]

OpenStudy (anonymous):

we are still not done with the other one but i am going to bet it is \(\frac{1}{8}\) as well

OpenStudy (anonymous):

probably easiest if we made a tree diagram have you done that?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

trying to recall

OpenStudy (anonymous):

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OpenStudy (anonymous):

there is my crappy picture starting the selecting a blue ball at first, which happens with probability \(\frac{1}{2}\) so the probability you end up at each of the 4 possibilities for the number of possible blue balls is \(\frac{1}{8}\) compute each probability, multiply by \(\frac{1}{8}\) and add

OpenStudy (anonymous):

if tails appears, two blue ones are places. so isnt it 7/10 ? 5+2 im not sure if the blue one will be drawn out and two will be put back?

OpenStudy (anonymous):

if that is what it means, yes i thought it meant it and another are replaced

OpenStudy (zarkon):

if the first one is blue then there is going to be either 9 or 11 balls in the bag...so shouldn't we get \[\frac{1}{2}\times \frac{4}{9}+\frac{1}{2}\times \frac{6}{11}\] for the second blue

OpenStudy (anonymous):

yeah i am messing this all up

OpenStudy (anonymous):

OpenStudy (zarkon):

lol

OpenStudy (zarkon):

you pull those up really fast ;)

OpenStudy (anonymous):

you said it probably best to make a tree diagram to begin with

OpenStudy (anonymous):

okay. i agree with the tree diagram

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