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ViennaJade

Approximately 15% of peole are left-handed. If two people are selected at random, what is the probability of the following events? P(At least one is right-handed)? Is this correct? 1-P(none are RH) so..1-P(LH and LH) 1-(.85)(.85) =1-.7225 =.2775...is this correct?

  • 10 months ago
  • 10 months ago

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  1. amistre64
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    should we assume that people can either be left or right handed, but not both?

    • 10 months ago
  2. ViennaJade
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    well I guess it's possible to be both

    • 10 months ago
  3. ViennaJade
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    but I've never had a question ask me that...

    • 10 months ago
  4. amistre64
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    since no stats are given as to ambidextrous peoples ... P(l) = .15, P(r) = .85 at least one is P(l) lr + ll

    • 10 months ago
  5. amistre64
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    at least one is right handed that is .... lr + rr

    • 10 months ago
  6. ViennaJade
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    Thank you! that makes sense.

    • 10 months ago
  7. amistre64
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    the possible ways to choose 2 people are: ll .15*.15 lr .15*.85 rl .85*.15 rr .85*.85 at least 1 r in it is: lr + rl + rr does that sound right?

    • 10 months ago
  8. amistre64
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    1 - ll :) .9775

    • 10 months ago
  9. ViennaJade
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    yes it does. makes so much more sense. Thank you!

    • 10 months ago
  10. ViennaJade
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    wait so is the answer .85?

    • 10 months ago
  11. amistre64
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    ll+lr+rl+rr = 1 lr+rl+rr = 1 - ll since l = .15 ... 1 - .15^2 = .9775

    • 10 months ago
  12. ViennaJade
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    oh sorry I forgot the one part..thank you!

    • 10 months ago
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