FUn questions - Probability - help

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The discussion centers on calculating the probability that John will make the Olympic team given that Bill finishes in the top three. The initial probabilities are that John has a 0.40 chance and Bill has a 0.25 chance of finishing in the top three, with Bill's probability increasing to 0.35 if John does not. The user seeks clarification on applying Bayes's Law to find the conditional probability P(J|B). The solution involves using the law of total probability to establish relationships between the probabilities and ultimately solve for the desired conditional probability. Understanding this calculation will help the user grasp similar probability questions in the future.
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FUn questions - Probability - help!

Hello! I need help understanding how to get to the answers on this question. I would be sooo appreciative if someone woule write it out and explain it, as once I understand this I can understand a lot more! I REALIZE they are VERY BASIC! so for many it will be easy, for me its not so easy!

Here we go!

IN an upcoming race, athletes have to finish first, second, or third in order to be selected for the team. The probabolity that John will finish in the top 3 is .40, wereas the probability that bill will finish in the top 3 is only .25. However, if John doesn't finish in the top 3, then Bills probabiltiy of finishing in the top 3 increases to .35. IF Bill was to finish in the top 3, what is the probability that Jogn will get on the olympic team?

The answer is .16.

Please how do we get there? I think we use the equation of P(A/B)= P(AnB)/ P(B)?

Thank u!
 
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P(J|B)P(B) = P(JnB) = P(BnJ) = P(B|J)P(J) .
 


Im still having a hard time, cud someone please put the numbers into the equation for me? Once I see this I will understand SO MANY more questions! THank you sooo mUCH!
 


IN an upcoming race, athletes have to finish first, second, or third in order to be selected for the team. The probabolity that John will finish in the top 3 is .40, wereas the probability that bill will finish in the top 3 is only .25. However, if John doesn't finish in the top 3, then Bills probabiltiy of finishing in the top 3 increases to .35. IF Bill was to finish in the top 3, what is the probability that Jogn will get on the olympic team?

P{J} = 0.4; P{B} = 0.25; P{B|~J} = 0.35; P{J|B} = ?

Law of total probability, or Bayes's Law:
P{B} = P{B|J}P{J} + P{B|~J}P{~J}
0.25 = P{B|J} 0.4 + 0.35 (1-0.4)
Solve for P{B|J}, call it x

From the definition of conditional prob.:
P{JnB} = P(BnJ) = P(B|J)P(J) = x P{J} = x 0.4
P{J|B} = P{JnB}/P{B} = x 0.4/0.25

EnumaElish
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I would definitely have logged in as EnumaElish had PF administration awarded that account the privilege of posting replies, after I reset my e-mail address Tuesday, October 28, 2008.
 


Thank u!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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