Solving a Probability Problem with Bayes Theorem

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To find the probability that the last ball drawn is red given that the first ball drawn is not white, Bayes' theorem and conditional probability can be applied. The problem involves calculating outcomes based on the remaining balls after the first draw. The initial setup includes 5 red, 5 black, and 5 white balls, leading to a complex probability tree. Simplifying the calculations and focusing on the relevant events can help in solving the problem. This approach will clarify the relationships between the events and lead to the desired probability.
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Homework Statement


you have a sack with 5 red balls, 5 black balls and 5 white balls. you take them out one at a time without returning until the sack is empty. what is the probability the last ball you took out is red given that the first ball taken out is not white.



Homework Equations


it should be solvable using the bayes theorom, the definition of conditioned probabilty, and the complete probability formula.



The Attempt at a Solution


i tried playing around with different events and using the formulas above but no luck... the main problem is that the probability tree is huge. i was hoping i can get a push in the right direction.

thanks.
 
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Hi ENgez! :smile:

You have a sack with 5 red balls, 4 black balls and 5 white balls. You take them out one at a time without returning until the sack is empty. what is the probability the last ball you took out is red? :wink:
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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