anonymous
  • anonymous
the results of a medical test show that of 32 people selected at random who were given the test, 2 tested positive and 30 tested negative. Determine the odds in favor of a person selected at random testing positive on the test.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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kropot72
  • kropot72
The odds are the ratio of the probability of an event occurring to that of its not occurring. What is the experimental probability of a randomly tested person testing positive?
anonymous
  • anonymous
you lost me with that last part.
kropot72
  • kropot72
The experimental probability of a randomly tested person testing positive is given by: \[\large \frac{number\ testing\ positive}{total\ number\ tested}\]

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anonymous
  • anonymous
so it would be as simple as 2/64?
anonymous
  • anonymous
then simplify it obviously
kropot72
  • kropot72
Where did '64' come from?
anonymous
  • anonymous
crap so it would just be "32" sorry insomnia is setting in!
kropot72
  • kropot72
Yes, the experimental probability of a randomly tested person testing positive is 2/32. Next step: What is experimental probability of a randomly tested person testing negative?
anonymous
  • anonymous
so it would be 30/32?
anonymous
  • anonymous
testing negative.
kropot72
  • kropot72
Correct. So looking at the definition of odds: 'The odds are the ratio of the probability of an event occurring to that of its not occurring.' So an initial result for the required odds in favor of a person selected at random testing positive on the test is: 2/32 : 30/32 which can be simplified. Can you simplify it?
kropot72
  • kropot72
The aim is to simplify \[\large \frac{2}{32}:\frac{30}{32}\] to get an integer on each side.
kropot72
  • kropot72
Multiply each term by 32/2
kropot72
  • kropot72
\[\large (\frac{2}{32}\times\frac{32}{2}):(\frac{30}{32}\times\frac{32}{2})=?\]
anonymous
  • anonymous
Sorry I was reading the book and it gave me a weird formula I was trying to wrap my head around based on what we were working on
anonymous
  • anonymous
it should be 1:15 if I did my math right
kropot72
  • kropot72
Correct :)
anonymous
  • anonymous
Theres a formula for odds in favor, the way you just walked out, is that the same process?
kropot72
  • kropot72
If the probability of an event A occurring is P(A) and the probability of event A not occurring is \[\large P(\bar{A})\] then the odds in favor of event A is given by \[\large P(A):P(\bar{A})\] This is the method that I used.
anonymous
  • anonymous
awesome, ok I think I have it. up for helping me with a couple more?
kropot72
  • kropot72
You're welcome :)

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