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Math Help - Counting, Probability, Trees, etc in Discrete Math

  1. #1
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    Counting, Probability, Trees, etc in Discrete Math

    Sorry for so many questions today guys. But this set will be last set for this week.

    Have some probability problems I'm doing in discrete math. I have done most of them but have trouble on starting for some of them and want to check answers on some as well.

    Here I go:


    Ok for problem 8:
    there are 36 possible outcomes
    e isn't a trick question is it? it's basically the same as a thru d correct?
    i think i have this question done correctly

    Problem 9: should be number of multiples of 7 plus multiples of 3 for 5 digits minus number of multiples for 21. so i think i have this done correctly

    Problem 10 is what I'm having trouble with. Seems like there are many ways total and I was going to use the probability tree but it doesn't work as well as it would on problem 11.

    Problem 11: used tree, got 88888 different pins.

    Haven't started 13..but should be similar to 12...no?

    14. I used a formula. P(n,r) = n(n-1)(n-2)...(n-r+1)
    i set the equation as P(n,5) = (n-5+1) then n-6=2027400
    got n as 2027406 and used that for (2027406)!/(5!*2027401!)


    Any input would greatly be appreciated, thanks for you help again!
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  2. #2
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    Ok for problem 8:
    there are 36 possible outcomes
    e isn't a trick question is it? it's basically the same as a thru d correct?
    Don't know if it is a trick question. My answer is 16/36.

    Problem 9: should be number of multiples of 7 plus multiples of 3 for 5 digits minus number of multiples for 21.
    I agree.

    Problem 10 is what I'm having trouble with. Seems like there are many ways total and I was going to use the probability tree but it doesn't work as well as it would on problem 11.
    I draw a binary tree in case A wins the first game. From there I got 10 possible developments. There is the same number of developments when B wins the first game.

    Problem 11: used tree, got 88888 different pins.
    8 variants for the first character, 10 variants for each of the rest, so 8 * 10^4 total.

    14. I used a formula. P(n,r) = n(n-1)(n-2)...(n-r+1)
    i set the equation as P(n,5) = (n-5+1) then n-6=2027400
    got n as 2027406 and used that for (2027406)!/(5!*2027401!)
    It makes no sense why from P(n,r) = n(n-1)(n-2)...(n-r+1) and r = 5 you concluded P(n,5) = (n-5+1). Where are the other four factors? Also, n - 5 + 1 is not n - 6.

    Note, however, that \displaystyle{n\choose r}=\frac{P(n,r)}{r!}.
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  3. #3
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    Hello, hellfire127!

    9. How many five-digit numbers are multiple of 3 or 7?

    There are 90,000 five-digit numbers.

    . . . . \left[\dfrac{900,\!000}{3}\right] \:=\:300,\!000 are multiples of 3.

    . . . . \left[\dfrac{90,\!000}{7}\right] \:=\:128,\!571 are multiples of 7.

    But: . \left[\dfrac{90,\!000}{21}\right] \:=\:42,\!857 are multiples of 3 and 7.
    . . . . . . . . . . . . . . . These have been counted twice.



    Therefore: . 300,\!000 + 128,\!571 - 42,\!857 \:=\:385,\!714

    . . . . . . . . . . are multiples of 3 or 7.

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, hellfire127!


    There are 90,000 five-digit numbers.

    . . . . \left[\dfrac{900,\!000}{3}\right] \:=\:300,\!000 are multiples of 3.

    . . . . \left[\dfrac{90,\!000}{7}\right] \:=\:128,\!571 are multiples of 7.

    But: . \left[\dfrac{90,\!000}{21}\right] \:=\:42,\!857 are multiples of 3 and 7.
    . . . . . . . . . . . . . . . These have been counted twice.



    Therefore: . 300,\!000 + 128,\!571 - 42,\!857 \:=\:385,\!714

    . . . . . . . . . . are multiples of 3 or 7.

    wait i got 38,572.

    i got 30000 that are multiples of 3 for 5 digits and 12857 multiples of 7 for 5 digits. i added these two and subtracted multiples of 21 for 5 digits which is 4285. 30000+12857-4285 = 38572

    or is this way for 5 digit numbers for multiples of 3 AND 7? does the wording of "Or" and "And" affect this type of problem??
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  5. #5
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    Quote Originally Posted by emakarov View Post
    Don't know if it is a trick question. My answer is 16/36.

    I agree.

    I draw a binary tree in case A wins the first game. From there I got 10 possible developments. There is the same number of developments when B wins the first game.

    8 variants for the first character, 10 variants for each of the rest, so 8 * 10^4 total.

    It makes no sense why from P(n,r) = n(n-1)(n-2)...(n-r+1) and r = 5 you concluded P(n,5) = (n-5+1). Where are the other four factors? Also, n - 5 + 1 is not n - 6.

    Note, however, that \displaystyle{n\choose r}=\frac{P(n,r)}{r!}.
    Is it just me because i only see 14 that total of the numbers on the faces is 8 or that at least one of the numbers i 6. counted all that add up to 8 and ones that contain at least a 6.

    thanks i'll use that binary tree and see if i can get an aswer.

    as for problem 11: for the pin has to be 5 characters total? i was thinking that the worded question means the pin could be 1 pin, 2 pin, 3 pin, 4 pin, 5 pin. i then chose 8 options under each and 10 there after, example:
    1-pin: 8 = 8
    2-pin: 8*10 = 80
    3-pin: 8*10*10 = 8x10^2
    4-pin: 8x10^3
    5 pin: 8x10^4
    i then added totals for each one. but if the pin is 5 total characters then yes I see how you got your answer. just the wording confused me a bit

    on the last problem, how would I go about solving for n?
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  6. #6
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    Hello agaibn, hellfire127!

    12. A poker hand consist of 5 cards drawn from a 52-card deck.
    How many different poker hands are possible?

    This is a "combination" since the order of the cards is not considered.

    Answer: . _{52}C_5} \:=\:\dfrac{52!}{5!\,47!} \;=\; 2,\!598,\!960\text{ possible poker hands.}




    13. In poker a hand is called "one pair" if and only if two of the cards
    are of the same value, and the other three cards have three different values,
    each differing from the pair.

    (a) In how many different ways can a poker hand be "one pair"?

    (b) What is the probability of "one pair"?

    This takes some Deep Thought . . .

    There are 13 choices for the value of the Pair.

    There are: . _4C_2 \:=\:6 ways to get two of the four cards of that value.


    For the other three (unmatched) cards, we have this routine . . .

    Card #1 can be any of the 48 cards which do not match the Pair.

    Card #2 can be any of the other 44 cards
    . . which do not match the Pair or card #1.

    Card #3 can be any of the other 40 cards
    . . which do not match the Pair or card #1 or card #2.

    But 48\cdot44\cdot40 imparts an order to the three cards.

    \text{To eliminate the ordering, divide by }3! :\;\;\dfrac{48\cdot44\cdot40}{6} \:=\:14,\!080

    Therefore, there are:

    . . 13 \times 6 \times 14,\!080 \:=\:1,098,240 poker hands with "one pair". . (a)



    P(\text{one pair}) \;=\;\dfrac{1,\!098,\!240}{2,\!598,\!960} \;=\;\dfrac{352}{833} .(b)

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  7. #7
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    Is it just me because i only see 14 that total of the numbers on the faces is 8 or that at least one of the numbers i 6. counted all that add up to 8 and ones that contain at least a 6.
    You are right; I counted (1, 7) and (7, 1) accidentally.

    For problem 11, you are right again. I missed "1 to 5" characters.

    on the last problem, how would I go about solving for n?
    You don't need n.
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  8. #8
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    Quote Originally Posted by Soroban View Post
    Hello agaibn, hellfire127!


    This is a "combination" since the order of the cards is not considered.

    Answer: . _{52}C_5} \:=\:\dfrac{52!}{5!\,47!} \;=\; 2,\!598,\!960\text{ possible poker hands.}





    This takes some Deep Thought . . .

    There are 13 choices for the value of the Pair.

    There are: . _4C_2 \:=\:6 ways to get two of the four cards of that value.


    For the other three (unmatched) cards, we have this routine . . .

    Card #1 can be any of the 48 cards which do not match the Pair.

    Card #2 can be any of the other 44 cards
    . . which do not match the Pair or card #1.

    Card #3 can be any of the other 40 cards
    . . which do not match the Pair or card #1 or card #2.

    But 48\cdot44\cdot40 imparts an order to the three cards.

    \text{To eliminate the ordering, divide by }3! :\;\;\dfrac{48\cdot44\cdot40}{6} \:=\:14,\!080

    Therefore, there are:

    . . 13 \times 6 \times 14,\!080 \:=\:1,098,240 poker hands with "one pair". . (a)



    P(\text{one pair}) \;=\;\dfrac{1,\!098,\!240}{2,\!598,\!960} \;=\;\dfrac{352}{833} .(b)

    yea 12 was easy so i didnt put up an answer. 13 is harder, thanks for the steps. i'll go over them again.
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  9. #9
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    ok so can anybody tell me which is the correct way for number 9?

    so problem 10 should be 20 possible outcomes correct? made a tree and came out to be 10 each team from the given method by emkarov.

    also problem 11 is the same way, don't know what the wording meant. if 5 character pins then i see that emkarov's solution makes sense.

    for problem 12, that was probably the easiest.

    13, thanks Soroban for the steps.

    as for last problem, i will have to look at it again.
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  10. #10
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    so problem 10 should be 20 possible outcomes correct?
    Yes.

    also problem 11 is the same way, don't know what the wording meant. if 5 character pins then i see that emkarov's solution makes sense.
    As I said, I think your first solution is right because it considers PINs of different lengths.
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  11. #11
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    Quote Originally Posted by emakarov View Post
    Yes.

    As I said, I think your first solution is right because it considers PINs of different lengths.
    ok yea sorry i must have overlooked that part of your post. thanks again for your help.

    now for the last few problems....
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  12. #12
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    problem 14 wouldn't be 2027400!/5!*2027395! would it?
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  13. #13
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    \displaystyle{n\choose 5}=\frac{P(n,5)}{5!}=\frac{2027400}{5!}
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  14. #14
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    Quote Originally Posted by emakarov View Post
    \displaystyle{n\choose 5}=\frac{P(n,5)}{5!}=\frac{2027400}{5!}
    Would this formula work for problems like \displaystyle{5\choose 3}[/QUOTE] and \displaystyle{8\choose 2}[/QUOTE] ? i'm used to solving them without variables.

    also, what do you think of problem 9 emakarov?
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  15. #15
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    Quote Originally Posted by hellfire127 View Post
    ok so can anybody tell me which is the correct way for number 9?
    \displaystyle\left\lfloor {\frac{{99999}}<br />
{3}} \right\rfloor  + \left\lfloor {\frac{{99999}}<br />
{7}} \right\rfloor  - \left\lfloor {\frac{{99999}}<br />
{{21}}} \right\rfloor  - \left( {\left\lfloor {\frac{{9999}}<br />
{3}} \right\rfloor  + \left\lfloor {\frac{{9999}}<br />
{7}} \right\rfloor  - \left\lfloor {\frac{{9999}}<br />
{{21}}} \right\rfloor } \right)=38572
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