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Math Help - Probability - not good with, can you help.

  1. #1
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    Probability - not good with, can you help.

    A bag contains five red candies, six white candies, and seven blue candies. Suppose one piece of candy is drawn at random. Find the probability of each of the following.
      • A white candy is drawn. _____________

      • A red or blue candy is draw. _____________

      • Neither a white nor a blue candy is drawn. _____________
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  2. #2
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    Quote Originally Posted by doodle08 View Post
    A bag contains five red candies, six white candies, and seven blue candies. Suppose one piece of candy is drawn at random. Find the probability of each of the following.

      • A white candy is drawn. _____________


      • A red or blue candy is draw. _____________


      • Neither a white nor a blue candy is drawn. _____________
    In each case, the required probability is equal to (number of favourable outcomes)/(total number of outcomes).

    eg. The answer to the first one is 6/18.
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  3. #3
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    Quote Originally Posted by doodle08 View Post
    A bag contains five red candies, six white candies, and seven blue candies. Suppose one piece of candy is drawn at random. Find the probability of each of the following.

      • A white candy is drawn. _____________




      • A red or blue candy is draw. _____________




      • Neither a white nor a blue candy is drawn. _____________


    You have a total of 18 candies.
    We'll assume that they're all equally likely to be chosen.
    Now, suppose I'm feeling mean, and I poison 1 of the 18 pieces of candy.
    So there's 18 candies, and 1 of them is poisonous.
    So if you pick 1, there's a 1 in 18 chance it will be the poisonous one - fair enough?
    So Probability( Candy = poisonous) = 1/18.

    Now, what if, instead, I poison two candies?
    Now there are 2 of the 18 candies which are poisonous.
    So if you pick one, you're twice as likely to get a poisonous one, so
    P( Candy = poisonous) = 2/18 = 1/9. Ok?

    Similarly if I poison 3 candies, the P(Candy = poisonous) = 3/18 = 1/6.

    Now, what if I poisoned all the white candies and not the other ones? (And didn't tell you that was what I was doing - because then you wouldn't eat the white ones, so my dastardly plan wouldn't work.)

    Well, there are 6 white candies. So what would be the probability of you picking out a poisonous/white one?




    ............

    For the second one, suppose I poisoned all the blue and red candies...

    ( I've been in the office too long... I think my psychotic tendencies are bubbling to the surface...)

    What would be the chances of you picking a poisonous/red or blue candy?

    .........

    For the last bit - suppose I don't poison the blue candies, and I don't poison the white candies. But I poison all the others, ie. the reds.

    What are the chances of you picking a poisonous one now? ie. NOT a white or blue one?
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  4. #4
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    Angry

    Quote Originally Posted by mr fantastic View Post
    In each case, the required probability is equal to (number of favourable outcomes)/(total number of outcomes).

    eg. The answer to the first one is 6/18.

    A pox on you, mr. fantastic. You have foiled my evil plan
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