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Math Help - Please help with simple probability

  1. #1
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    Please help with simple probability

    Hello, I have a few simple questions that I need help with. I have just started learning this.

    1) A survey of 200 hundred teenagers showed that during the Christmas holidays, 120 watched cricket regularly, 85 watched tennis regularly and 60 were not regular viewers of either sport. What is the probability that a teenager selected at random from the group regularly watched:

    a) both cricket and tennis
    b) cricket but not tennis
    c) at least on of the two sports

    We are given this formula: P(AUB) = P(A) + P(B) - P(A^B) **where ^ is intersection).

    How do I determine the intersection of A and B? Could someone please show me how to set these out in proper set notation?

    2) A box contains two red and five blue pens. Two pens are selected at random. What is the probability that the second pen is blue?

    We are given the conditional formula: P(B|A) = P(A^B)/P(A)

    Again, if someone could show me the steps for setting out in the notation it would be very much appreciated. I just need an example to get my head around.

    Thanks.
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  2. #2
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    Quote Originally Posted by Chekhov View Post
    Hello, I have a few simple questions that I need help with. I have just started learning this.

    1) A survey of 200 hundred teenagers showed that during the Christmas holidays, 120 watched cricket regularly, 85 watched tennis regularly and 60 were not regular viewers of either sport. What is the probability that a teenager selected at random from the group regularly watched:

    a) both cricket and tennis
    b) cricket but not tennis
    c) at least on of the two sports

    We are given this formula: P(AUB) = P(A) + P(B) - P(A^B) **where ^ is intersection).

    How do I determine the intersection of A and B? Could someone please show me how to set these out in proper set notation?

    Mr F says: I'd suggest you first try drawing a Venn diagram.

    2) A box contains two red and five blue pens. Two pens are selected at random. What is the probability that the second pen is blue?

    We are given the conditional formula: P(B|A) = P(A^B)/P(A)

    Again, if someone could show me the steps for setting out in the notation it would be very much appreciated. I just need an example to get my head around.

    Mr F says: I'd suggest you first try drawing a tree diagram.

    Thanks.
    ..
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  3. #3
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    Mr. F,

    I have tried drawing a Venn diagram... but what I can't seem to grasp is how to determine the intersection of the two events. Am i doing this the wrong way? Either way I'm missing something; it's only my first day on this subject.

    Thanks.
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  4. #4
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    Quote Originally Posted by Chekhov View Post
    Hello, I have a few simple questions that I need help with. I have just started learning this.

    1) A survey of 200 hundred teenagers showed that during the Christmas holidays, 120 watched cricket regularly, 85 watched tennis regularly and 60 were not regular viewers of either sport. What is the probability that a teenager selected at random from the group regularly watched:

    a) both cricket and tennis
    b) cricket but not tennis
    c) at least on of the two sports

    We are given this formula: P(AUB) = P(A) + P(B) - P(A^B) **where ^ is intersection).

    How do I determine the intersection of A and B? Could someone please show me how to set these out in proper set notation?
    Disclaimer: This is NOT my strongest subject but I will try...

    define the following events

    C = \mbox{watch cricket} \\\ C^c=\mbox{don't watch cricket}
    T=\mbox{watch tennis} \\\ T^c=\mbox{ don't watch tennis}
    So here is what we know

    P(C)=\frac{120}{200} \\\ P(T)=\frac{85}{200} \\\ P[(C \cup T)^c]=\frac{60}{200}

    We know that

    P[(C \cup T)]+P[(C \cup T)^c]=1 \iff P[(C \cup T)]=1-P[(C \cup T)^c]=1-\frac{60}{200}=\frac{140}{200}

    P(C \cup T)=P(C)+P(T)-P(C \cap T) \iff \frac{140}{200}=\frac{120}{200}+\frac{85}{200}-P(C \cap T)

    P(C \cap T)=\frac{65}{200}

    I think that this should work for the rest aswell.

    Remember that P(A) + P(A^c)=1 That is the probability of A plus the probabilty of its compliment equal 1.

    Good luck.
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  5. #5
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    Quote Originally Posted by Chekhov View Post
    Mr. F,

    I have tried drawing a Venn diagram... but what I can't seem to grasp is how to determine the intersection of the two events. Am i doing this the wrong way? Either way I'm missing something; it's only my first day on this subject.

    Thanks.
    Let x be the number that watch both tennis and cricket (the intersection). Then the number watching cricket only is 120 - x and the number watching tennis only is 85 - x.

    Note that the number watching tennis or cricket is 200 - 60 = 140.

    Then (120 - x) + x + (85 - x) = 140 => x = 65 (which is confirmed by TheEmptySet's analysis, by the way).

    Now you should be able to put the numbers into the Venn diagram and answer the questions.
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  6. #6
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    Thank you both for your help. TheEmptySet: The setting out you detailed was helpful and did allow me to solve the other problems. Mr. Fantastic: the suggestion about the tree diagram was helpful--as soon as I drew it I could see how to solve the problems. I've got so much to learn!
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