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  1. #1
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    Questions Help

    I've got two questions:
    1. A mother knows that 20% of children who accept invitation to birthday parties do not come. If she invites 12 children to a party and only has 10 party hats, what is the probability that there is not a hat for every child who comes?
    The mother knows that there is a probability of 0.1 that a child who comes refuses to wear a hat. if this is taken into account, what is the probability that the number of hats will not be adequate?

    I know that the probability of a child not coming is \frac{1}{5}. So the probability that 11 or 12 come will be
    P(not enough hats)=(\frac{4}{5})^{11}(\frac{1}{5})+(\frac{4}{5}  )^{12}=0.0859 the answer for this part is supposed to be 0.28, without the first part i don't see that the second part can be done.

    2.When a boy fires a rifle at a range the probability that he hits the target is p.
    (a) Find the probability that, firing 5 shots, he scores at least 4 hits.
    (b) Find the probability that, firing n shots (n\geq2), he scores at least two hits.
    For (a), there would be the probability of hitting 4 or 5 times. p^4(1-p)+p^5=p^4 which is not the answer. Answer is [tex]5p^4-4p^5[\math]
    (b) answer would be 1 minus the probability of missing all and hitting 1.
    [tex]1-p(1-p)^{n-1}-(1-p)^n[\math]. answer is 1-np(1-p)^{n-1}-(1-p)^n.

    For all of these where have i got wrong?
    Thanks for any help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arze View Post
    I've got two questions:
    1. A mother knows that 20% of children who accept invitation to birthday parties do not come. If she invites 12 children to a party and only has 10 party hats, what is the probability that there is not a hat for every child who comes?
    The mother knows that there is a probability of 0.1 that a child who comes refuses to wear a hat. if this is taken into account, what is the probability that the number of hats will not be adequate?

    I know that the probability of a child not coming is \frac{1}{5}. So the probability that 11 or 12 come will be
    P(not enough hats)=(\frac{4}{5})^{11}(\frac{1}{5})+(\frac{4}{5}  )^{12}=0.0859
    The probability that 11 or 12 come is:

    P(\text{not enough hats})=12 \times 0.8^{11}\times 0.2+0.8^{12}\approx 0.275

    CB
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  3. #3
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    Quote Originally Posted by arze View Post
    I've got two questions:
    1. A mother knows that 20% of children who accept invitation to birthday parties do not come. If she invites 12 children to a party and only has 10 party hats, what is the probability that there is not a hat for every child who comes?
    The mother knows that there is a probability of 0.1 that a child who comes refuses to wear a hat. if this is taken into account, what is the probability that the number of hats will not be adequate?

    I know that the probability of a child not coming is \frac{1}{5}. So the probability that 11 or 12 come will be
    P(not enough hats)=(\frac{4}{5})^{11}(\frac{1}{5})+(\frac{4}{5}  )^{12}=0.0859 the answer for this part is supposed to be 0.28, without the first part i don't see that the second part can be done.
    Or,

    there is a \frac{4}{5} probability of a child coming and a \frac{10}{12} probability of a hat being available and a \frac{9}{10} probabilty of deciding to wear it.

    The probability of a hat being worn by a child is \frac{4}{5}\frac{10}{12}\frac{9}{10}

    This means the "statistical" fractional number of hats out of 10 being worn is 12 times that (it's just a way of thinking of it! as we can't have a fraction of a hat).

    As there are 10 hats available, divide by 10 for the probability..

    The probability of all 10 hats being worn is \frac{12}{10}\frac{4}{5}\frac{10}{12}\frac{9}{10}=  \frac{36}{50}=\frac{72}{100}=0.72

    Therefore the probability a hat is not available to a child is 0.28
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    Or,

    there is a \frac{4}{5} probability of a child coming and a \frac{10}{12} probability of a hat being available and a \frac{9}{10} probabilty of deciding to wear it.

    The probability of a hat being worn by a child is \frac{4}{5}\frac{10}{12}\frac{9}{10}

    This means the "statistical" fractional number of hats out of 10 being worn is 12 times that (it's just a way of thinking of it! as we can't have a fraction of a hat).

    As there are 10 hats available, divide by 10 for the probability..

    The probability of all 10 hats being worn is \frac{12}{10}\frac{4}{5}\frac{10}{12}\frac{9}{10}=  \frac{36}{50}=\frac{72}{100}=0.72

    Therefore the probability a hat is not available to a child is 0.28
    There are two parts and two answers to this question. The first part does not include the probability that a child refuses to wear the hat, and the answer is 0.28. when the probability that a child refuses to wear a hat is considered, the answer is 0.11.
    Thanks
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Archie Meade View Post
    Or,

    there is a \frac{4}{5} probability of a child coming and a \frac{10}{12} probability of a hat being available and a \frac{9}{10} probabilty of deciding to wear it.

    The probability of a hat being worn by a child is \frac{4}{5}\frac{10}{12}\frac{9}{10}

    This means the "statistical" fractional number of hats out of 10 being worn is 12 times that (it's just a way of thinking of it! as we can't have a fraction of a hat).

    As there are 10 hats available, divide by 10 for the probability..

    The probability of all 10 hats being worn is \frac{12}{10}\frac{4}{5}\frac{10}{12}\frac{9}{10}=  \frac{36}{50}=\frac{72}{100}=0.72

    Therefore the probability a hat is not available to a child is 0.28
    Pardon?!

    CB
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  6. #6
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    Sincere apologies lads!
    I saw 0.28 and didn't even think it through.
    Red card offense, a spell on the sidelines would do me good...

    If 11 or 12 children come, then there will be a shortage of hats.
    Any of 12 children can decide not to come.
    The probability of only one particular child not coming is \frac{1}{5}\ (\frac{4}{5})^{12}

    Since any of 12 children can choose not to arrive

    P(child\ missing)=\frac{12}{5}\ (\frac{4}{5})^{11}

    If all 12 come there will be a shortage,
    this can happen in one way only.

    Hence, P(shortage)=\frac{12}{5}\ (\frac{4}{5})^{11}+(\frac{4}{5})^{12}=0.275

    Part (ii)

    There is a 90% chance of a child deciding to choose a hat on arriving.

    Hence, if 11 come and all decide to choose,
    or, if 12 come and 11 or 12 choose a hat,
    there will be a shortage.

    P(shortage)=\frac{12}{5}(\frac{4}{5})^{11}(0.9)^{1  1}+(\frac{4}{5})^{12}[12(0.9)^{11}(0.1)+(0.9)^{12}]=0.105

    The above was edited as a factor was inadvertently omitted.
    Last edited by Archie Meade; January 28th 2010 at 09:36 AM. Reason: left out a factor
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  7. #7
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    corrections added above.
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  8. #8
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    Quote Originally Posted by arze View Post

    2.When a boy fires a rifle at a range the probability that he hits the target is p.
    (a) Find the probability that, firing 5 shots, he scores at least 4 hits.
    (b) Find the probability that, firing n shots (n\geq2), he scores at least two hits.
    For (a), there would be the probability of hitting 4 or 5 times. p^4(1-p)+p^5=p^4 which is not the answer. Answer is [tex]5p^4-4p^5[\math]
    (b) answer would be 1 minus the probability of missing all and hitting 1.
    [tex]1-p(1-p)^{n-1}-(1-p)^n[\math]. answer is 1-np(1-p)^{n-1}-(1-p)^n.

    For all of these where have i got wrong?
    Thanks for any help!
    The only thing you are missing is "counting the number of ways the event can happen".

    Q2 (a) There are \binom{5}{4} ways to miss once..

    on the 1st, 2nd, 3rd, 4th or 5th shot.

    P(miss\ once)=\binom{5}{4}p^4(1-p)=5p^4(1-p)=5p^4-5p^5

    P(at\ least\ one\ hit)=1-(5p^4-5p^5+p^5)=1-5p^4+4p^5

    Q2 (b) There are \binom{n}{n-1} ways to miss n-1 times, therefore to hit once.

    P(one\ hit)=\binom{n}{n-1}p(1-p)^{n-1}=np(1-p)^{n-1}

    P(no\ hit)=(1-p)^n

    P(at\ least\ 2\ hits)=1-(np[1-p]^{n-1}+[1-p]^n)
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    Part (ii)

    There is a 90% chance of a child deciding to choose a hat on arriving.

    Hence, if 11 come and all decide to choose,
    or, if 12 come and 11 or 12 choose a hat,
    there will be a shortage.

    P(shortage)=\frac{12}{5}(\frac{4}{5})^{11}+(\frac{  4}{5})^{12}[(0.9)^{11}+(0.9)^{12}]=0.105
    Can't get the last calculation you did. does not look correct either.
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  10. #10
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    Hi arze,

    in this case, if 12 children came and 2 of them decided not to wear a hat, then the number of hats will be adequate, since if any of the other children want one, there will be enough.

    There will be a shortfall only if 12 come and 12 or 11 want to wear a hat.

    If 11 come, there will be a shortfall only if they all want a hat.

    If 10 or less come, there is never a shortfall.

    The calculations reflect 11 children arriving and all choosing to wear a hat given that the probability of wanting one is 0.9.

    12 children arriving for which there is 2 cases...
    (a) all 12 want to wear a hat
    (b) 11 of the 12 want to wear a hat.

    Only in those 3 cases would a shortfall occur.
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  11. #11
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    I've tried that equation you gave above and it still does not get the answer 0.11
    I get the answer 0.247
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  12. #12
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    Sorry arze,

    I dropped the 0.9^{11} factor of \frac{12}{5}(\frac{4}{5})^{11}0.9^{11}

    apologies again..
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  13. #13
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    Thank you very much for all the help!
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  14. #14
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    Hi arze,
    i actually omitted 2 factors..

    P(shortfall)=\binom{12}{11}(0.2)^1(\frac{4}{5})^{1  1}(0.9)^{11}+(\frac{4}{5})^{12}(0.9)^{12}+(\frac{4  }{5})^{12}[\binom{12}{11}(0.9)^{11}(0.1)^1]

    P(shortfall)=\frac{12}{5}(\frac{4}{5})^{11}(0.9)^{  11}+(\frac{4}{5})^{12}[(0.9)^{12}+(1.2)(0.9)^{11}]=0.10998
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