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Thread: probably the long way round...

  1. #1
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    probably the long way round...

    Hi guys,

    the only way I can see to solve this probability question is rather manual. It may be the only way, but I suspect someone here will have a smarter solution. So, here goes:

    3 people independently think of an integer in the set {1,2,3,4,5,6,7}. Find in fractional form the probability that:
    a) all 3 of the integers selected are greater than 4
    b) the least integer selected is 5
    c) the 3 integers selected are different, given that the least integer selected is 5
    d) the sum of the 3 integers selected is more than 15

    so this is a problem with replacement.

    a) The probability of selecting a number greater than 4 is 3/7 and there are three numbers to select so the probability is $(3/7)^3 = \frac{27}{343}$

    b) if the least number selected is 5, there are 3 numbers 5, 6 and 7 that can be arranged in 27 ways. We require only those selections that contain a 5.
    so we have:

    5 5 5 x
    5 5 6 x
    5 5 7 x
    5 6 5 x
    5 6 6 x
    5 6 7 x y
    5 7 5 x
    5 7 6 x y
    5 7 7 x
    6 5 5 x
    6 5 6 x
    6 5 7 x y
    6 6 5 x
    6 6 6
    6 6 7
    6 7 5 x y
    6 7 6
    6 7 7
    7 5 5 x
    7 5 6 x y
    7 5 7 x
    7 6 5 x y
    7 6 6
    7 6 7
    7 7 5 x
    7 7 6
    7 7 7

    an x in column 4 identifies all the entries containing a 5. There are 19 'x's, so the Probability is $\frac{19}{343}$

    c) now we have to fish out the number of cases where all 3 numbers are different. Again the table is handy and I wouldn't know how to proceed without it.
    These are marked with a y in Column 5 and there are 6 values.

    d) to get the sum of the integers more than 15, it is just a case of writing out all the possible numbers and counting the cases. I won't write it all out here.
    As I said, my suspicion is that there its a better way!
    Last edited by s_ingram; Jun 8th 2017 at 05:09 AM.
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  2. #2
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    Re: probably the long way round...

    Quote Originally Posted by s_ingram View Post
    3 people independently think of an integer in the set {1,2,3,4,5,6,7}. Find in fractional form the probability that:
    a) all 3 of the integers selected are greater than 4
    b) the least integer selected is 5
    c) the 3 integers selected are different, given that the least integer selected is 5
    d) the sum of the 3 integers selected is more than 15
    so this is a problem with replacement.
    a) The probability of selecting a number greater than 4 is 3/7 and there are three numbers to select so the probability is $(3/7)^3 = \frac{27}{343}$
    b) if the least number selected is 5, there are 3 numbers 5, 6 and 7 that can be arranged in 27 ways. We require only those selections that contain a 5.
    an x in column 4 identifies all the entries containing a 5. There are 19 'x's, so the Probability is $\frac{19}{343}$

    c) now we have to fish out the number of cases where all 3 numbers are different. Again the table is handy and I wouldn't know how to proceed without it.
    These are marked with a y in Column 5 and there are 6 values.

    d) to get the sum of the integers more than 15, it is just a case of writing out all the possible numbers and counting the cases. I won't write it all out here.
    As I said, my suspicion is that there its a better way!
    You did both #a & #b correctly.
    However there is an easier way for #b.
    There $2^3=8$ ways with no fives. Then $3^3-2^3=19$.

    c) You have listed the space for c. There are only $27$ How many of those contain a five and are all different?
    Last edited by Plato; Jun 8th 2017 at 06:25 AM.
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  3. #3
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    Re: probably the long way round...

    Yes! I see exactly what you mean - and it is much quicker!
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  4. #4
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    Re: probably the long way round...

    For b), if the least integer selected is 5, then suppose the first person selects 5. Then the 2nd and 3rd person may select 5,6,or 7. Suppose the first person does not select 5, but the second person does. Then, the first person may select 6 or 7, the second person selects 5, and the 3rd person may select 5,6, or 7. Finally, suppose the first two people do not select 5, but the third person does. Then the first two people may select 6 or 7 while the third person selects 5. These are disjoint cases, so their probabilities add:

    $\dfrac{1}{7}\left( \dfrac{3}{7} \right)^2 + \dfrac{2}{7} \dfrac{1}{7} \dfrac{3}{7} + \left( \dfrac{2}{7} \right)^2 \dfrac{1}{7} = \dfrac{9+6+4}{343} = \dfrac{19}{343}$

    c) if the three numbers are different, then the numbers 5,6,7 are chosen in some order. The number of ways for the three people to choose three distinct numbers is $3!$, the number of permutations. Each permutation is disjoint (it is not possible for two permutations to be selected at the same time), so we may add the probabilities:

    $3!\left(\dfrac{1}{7}\right)^3 = \dfrac{6}{343}$

    d) How many ways are there to get a sum of exactly 16?
    We can have:
    2,7,7 - 3 orders
    3,6,7 - 3! orders
    4,5,7 - 3! orders
    4,6,6 - 3 orders
    5,5,6 - 3 orders
    21 different ways to get exactly 16
    How about exactly 17?
    3,7,7 - 3 orders
    4,6,7 - 3! orders
    5,6,6 - 3 orders
    5,5,7 - 3 orders
    15 different ways to get exactly 17
    For 18, there are 10 ways
    For 19, there are 6 ways
    For 20, there are 3 ways
    For 21, there is 1 way.

    These are triangle numbers. If you can show that the pattern (down to exactly 12) is always the next triangle number, this becomes easier to solve.

    Then, you just have $\dfrac{\displaystyle \sum_{n=1}^6\dfrac{n(n+1)}{2}}{343} = \dfrac{8}{49}$
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    Re: probably the long way round...

    Quote Originally Posted by s_ingram View Post
    3 people independently think of an integer in the set {1,2,3,4,5,6,7}. Find in fractional form the probability that:
    c) the 3 integers selected are different, given that the least integer selected is $5$
    Quote Originally Posted by SlipEternal View Post
    c) if the three numbers are different, then the numbers 5,6,7 are chosen in some order. The number of ways for the three people to choose three distinct numbers is $3!$, the number of permutations. Each permutation is disjoint (it is not possible for two permutations to be selected at the same time), so we may add the probabilities:
    $3!\left(\dfrac{1}{7}\right)^3 = \dfrac{6}{343}$
    @SlipEternal, You missed the conditional: given that the least integer selected is $5$. That cuts the size down to $27$.
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  6. #6
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    Re: probably the long way round...

    Quote Originally Posted by Plato View Post
    @SlipEternal, You missed the conditional: given that the least integer selected is $5$. That cuts the size down to $27$.
    Very true, thanks
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  7. #7
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    Re: probably the long way round...

    Quote Originally Posted by s_ingram View Post
    3 people independently think of an integer in the set {1,2,3,4,5,6,7}. Find in fractional form the probability that:
    d) the sum of the 3 integers selected is more than 15
    As I said, my suspicion is that there its a better way!
    The condition more than 15 makes me think that this is a counting question. I would approach by counting the positive cases. Look are the multi-subsets of three.
    No such set 'subset' may contain a $1$. WHY?
    Collections containing a $2$: $<2,7,7>[3]$ three ways. [number of ways to rearrange]
    Collections containing a $3$: $<3,7,7>[3],<3,6,7>[6]$ nine ways.
    Collections containing a $4$: $<4,7,7>[3],<4,6,6>[3],<4,6,7>[6],<4,5,7>[6]$ eighteen ways.
    There are twenty-seven, $3^3$ ordered triples containing $5's,~6's,\text{ or }7's$ only one of which, $<5,5,5>$, does not meet the condition. So $56$ ways to have the three add to more than $15$.
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