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Q) At a javalin competition two athletes (1 & 2) are competing against each other. Each has one attempt to throw the javalin. Assume the acheived distance of a throw ($L$1 & $L2$) [note these are processes] are exponentially distributed random variables and the preformance of each athlete is not affected by their competotors preformance. The average throwing distance for athletes 1 & 2 are $L1$ & $L2$ respectivley [note these are just some constant number, i think mean distances we have to calculate somewhere, not the same as the processes mentioned earlier].


i) Explicit expressions for the probabilities $P(Li = l)$ := $p(l)$ for each athlete to throw a javalin at distance l?

ii) The probability $P(L$1 - $L$2 = $L)$ := $P(L)$ that the result of athlete 1 is
different to the result of athlete 2 by $L$?

iii) The probability that athlete 1 will beat athlete 2?

iv) The probability that the world record $l$0 will not be beaten at this competition?

v) The probability that the winner of the event will beat the world record?

As far as the question goes i know i have to use the law of total probability somewhere to begin with as i have done another practise question, however that question was much easier to understand and im stuggeling with what i have to do exacly with this question, the wording has stumpted me. Moreover i have no idea with the final questions as their is no mention of world records in the Sub text. I would really appreciate help in how to tackle and solve problems like this to aid my preperation of my upcoming exam. Finally the bits in square brackets is just to clarify further from my understanding of the question, as it may not be obvious due to the way i have enetered the question. Thanks in advance

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Under a continuous distribution model, the probability of hitting a distance of exactly $l$ is $0$. And the probability that the two results will be different is $1$. The world record stuff makes sense. If I read your post correctly, the world record is specified as being $l_0$. Problem iii) is more difficult than the others. – André Nicolas Nov 19 '12 at 5:43
Thanks, but from my understanding its has alot more to it than simple probabilities. I know i have to form certain expressions/equations and then solve and simplfy. I dont think they are simple numerical solutions. And yes i do now see, that does seem to be the world record. – user49032 Nov 19 '12 at 9:49

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