SOLUTION: Assume that the pointer can never lie on a border line. Find the following probabilities. P(A), P(B), P(C), = 1st Wheel has 3 sections named A,B,C

SOLUTION: Assume that the pointer can never lie on a border line. Find the following probabilities. P(A), P(B), P(C), = 1st Wheel has 3 sections named A,B,C > I answered correctly 1/3 for

Question 579361: Assume that the pointer can never lie on a border line. Find the following probabilities.
P(A), P(B), P(C), = 1st Wheel has 3 sections named A,B,C > I answered correctly 1/3 for each of those sections in the first wheel.-
P(D), P(E), P(F), = 2nd Wheel has 3 sections named D,E,F
P(G), P(H), P(I) = 3rd Wheel has 4 sections named G,H.I,J J was not included in the question to account for.
Since there are 3 wheels, I assumed each wheel was divided into sections, logically I presumed 3 part wheel, divided into thirds and so on.
I was correct for the first wheel and incorrect for the next two wheels. I initially thought thirds as well for the second wheel and fourths for the third wheel.
I know I am over thinking this, but cannot seem to determine the correct fractional makeup for the second and third wheels. Can someone help me understand this?
Thank you
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
what was the exact wording of the problem?
do the 3 wheels spin independently of each other?
is this a joint probability, i.e. p(A) * p(D) * p(G)?
your assumption should be correct if p(D) was only from the spin of the second wheel.
same goes for p(G).
p(A) would be 1/3
p(D) would be 1/3
p(G) would be 1/4
this is obviously not the case so there must be something to the problem that we're missing.
send me the complete wording of the problem and also the solutions that are expected and i might be able to piece it together.

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