Solve the following? Stacy is playing with her magical colored wands. They come in three colors: red, yellow, and blue. Every hour, the wands multiply and change color with the following probabilities: (Continued in details)

#1.# A red wand has probability #5/12# of becoming two red wands, #1/3# of becoming one red wand and one yellow wand, and #1/4# of becoming one blue wand.

#2.# A yellow wand has probability #1/2# of becoming two yellow wands, #1/4# of becoming one yellow wand, and #1/4# of becoming one blue wand.

#3.# A blue wand never changes color.

If Alex starts with one red yarn ball, what is the probability that eventually all wands will be blue?

P.S. Please explain the answer as detailed as possible. Also, the information is exactly what the problem provides with.(Nothing added or removed)

1 Answer
MattyMatty
Feb 21, 2018

#1 - 0.2 sqrt(10) = 0.367544#

Explanation:

#"Name"#
#P[R] = " Probability that one R wand turns blue eventually"#
#P[Y] = " Prob. that one Y wand turns blue eventually."#
#P["RY"] = " Prob. that a R & Y wand both turn blue event."#
#P["RR"] = " Probability that two R wands turn blue event."#
#P["YY"] = " Probability that two Y wands turn blue event."#

#"Then we have"#
#P["RY"] = P[R]*P[Y]#
#P["RR"] = (P[R])^2#
#P["YY"] = (P[Y])^2#

#"So we get two equations in two variables P[R] and P[Y] :"#
#P[Y] = 1/4 + (1/4) P[Y] + (1/2) P[Y]^2#
#=> 2 P[Y]^2 - 3 P[Y] + 1 = 0#
#=> P[Y] = 1/2 " OR " cancel(1)#
#=> P[Y] = 1/2 " (P[Y] = 1 is not possible)"#
#P[R] = 1/4 + (1/3) P["RY"] + (5/12) P["RR"]#
#= 1/4 + (1/3)(1/2) P[R] + (5/12) P[R]^2#
#=> 5 P[R]^2 - 10 P[R] + 3 = 0#
#=> P[R] = (10 pm sqrt(40))/10#
#= 1 pm 2 sqrt(10)/10#
#= 1 - 0.2 sqrt(10)#
#= 0.367544#

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