Basic Probability Stuff

Discussion in 'Physics & Math' started by kingwinner, Apr 16, 2006.

  1. kingwinner Registered Senior Member

    Messages:
    796
    1) Determine the probability of rolling a composite number with one die.
    [The answer provided in my textbook is 2/9, and I am puzzled...shouldn't the correct answer be 1/3? Why is it 2/9?]

    2) A field-hockey team played 7 games and won 4 of them. There were no ties. How many arrangements are possible if the team have at least 2 wins in a row?
    [Should I calculate the complement for this question? If so, how can I find out that?]

    3) In how many ways can all the letters of the word CANADA be arranged if the consonants must always be in the order in which they occur in the word itself?
    [I tried to list them all out, and there are 20. But, is there a faster way to solve this problem without listing out everything?]

    Thank you for your help!

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  3. DaleSpam TANSTAAFL Registered Senior Member

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    Who picked out your textbook? I think it has been wrong at least once on each assignment you have had so far! Usually I am not a proponent of book-burning, but in this case I could make an exception.

    Out of the integers from 1 to 6 the only two composite numbers are 4 and 6. So the probability is 2/6 or 1/3. You are correct.


    The first thing to notice is that there is only 1 possible arrangement where you do not have 2 wins in a row (W L W L W L W). So all you have to do is determine the total number of arrangements (7 choose 4) and subtract 1.


    Once the location of the A's are fixed the consonant's locations are also fixed since the consonants are not changing order. And since the A's are interchangable their ordering is unimportant, only their locations. So basically you have only to determine how many ways you can choose 3 locations for A's out of 6 possible locations. As you correctly found, 6 choose 3 is 20.

    -Dale
     
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  5. kingwinner Registered Senior Member

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    796
    1) Oh, the answer in the textbook is wrong again! :bugeye: This is driving my crazy...

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    My school picked the textbook, I don't have any choice. There are way too many mistakes in my textbook...
     
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  7. kingwinner Registered Senior Member

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    796
    4) I have a question for those of you who know sports!

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    Are "ties" allowed in basketball, baseball, and hockey games, or only a "win" or a "loss" is possible? Are "ties" possible? I need this information for the next 2 questions.



    5) Suppose your school's basketball team is playing a four-game series against another school. So far this season, each team has won 3 of the 6 gemas in which they faced each other. Determine the probability of your school winning exactly 2 games.
    [Is a "tie" possible in one (or more) of the 4 games?]


    6) Suppose that the Toronto Blue Jays (baseball team, I guess) face the New York Yankees in the division final. In this best-of-five series, the winner is the first team to win 3 games. The games are played in Toronto and in New York, with Toronto hosting the first, second, and if needed, fifth geams. The consensus among experts is that Toronto has a 65% chance of winning at home and a 40% chance of winning in New York. What is the chance of Toronto winning in three straight games?
    [This seems to be a very complex probability problem. Outcomes seems not to be equally likely...can someone please explain how I can calculate the probability?]


    7) What is the probability of rolling a 7 and 2 with a die?
     
  8. mathman Valued Senior Member

    Messages:
    2,002
    4) In basketball and baseball, under ordinary circumstances there are no ties. However, baseball is played outdoors, so if there is a weather interuption or something like that (power failure), a tie is possible. Before there was night baseball, it happened occasionally. In hockey during the regular season, ties are possible, but rare. During the playoffs, the game will go on until a decision is reached.

    5) Straightforward combinatorial problem. Binomial theorem says there are respectively, 1,4,6,4,1 ways to win 0,1,2,3,4 games. Assuming (based on previous history) a .5 probability of winning any game, this gives you 3/8 as the probability of winning exactly 2 games.

    6) Three possibilities for Toronto, WWW, LWWW, LLWWW. You should be able to do the calculation for each case. Since they are mutually exclusive, you just have to add them up.

    7) Confusing question (a die means one die) - did you really mean a pair of dice? If so the relative frequencies of 2 through 12 are 1,2,3,4,5,6,5,4,3,2,1, so chance of 7 is 6/36 while chance of 2 is 1/36.
     
  9. kingwinner Registered Senior Member

    Messages:
    796
    6) I think by saying winning 3 straight games, only WWW is counted. (i.e. New York winning no games). But how can I calculate the probability for this?

    7) Let's skip this question...the wording of the question is very poor.
     
  10. shmoe Registred User Registered Senior Member

    Messages:
    524
    This is beside the main point, but ties are no longer possible in regular season hockey (NHL). They have a brief 4vs4 5minute sudden death overtime, then a shootout if necessary. The loser gets an "Overtime loss" and 1 point, the winner a regular "win" and 2 points.


    It's probably safe to assume that 5) and 6) are not allowing the possibility of a tie. It doesn't actually matter though- both are asking for certain amounts of wins and it doesn't matter if the non-win games are losses, ties, forfeits, or whatever, just that they aren't wins.

    For 6), I interpreted "winning in three straight games" as a WWW only. This is what the phrase usually means in my experience, but it could go either way I suppose. In any case, what is the probability for the first W? the second? the third? Just look at where the games are played.

    For 7), if we take "die" to mean a normal 6 sided die, then the probability of rolling a 7 is just 0.
     
  11. kingwinner Registered Senior Member

    Messages:
    796
    6) The question says "...In this best-of-five series, the winner is the first team to win 3 games...", does that mean if one of the 2 teams won 3 games, then no games will be played after that, or will 5 games be played anyway?
     
  12. shmoe Registred User Registered Senior Member

    Messages:
    524
    They stop playing once one of the teams wins 3.
     
  13. kingwinner Registered Senior Member

    Messages:
    796
    A question about odds...

    8) The odds in favor of the Titans winning the hockey tournment are 5:2. What would you win on a $20 wager if the Titans win?

    Solution:
    odds in favor of A=amount wagered / amount won
    5/2 = 20/x
    x=8
    Thus, you would win $8.

    I don't understand how they got the top formula "odds in favor of A=amount wagered / amount won", can someone explain this? Is the $20 wagered on the Titans winning or on the opposite team?
    Thanks a lot!
     
  14. przyk squishy Valued Senior Member

    Messages:
    3,203
    Everything you wanted to know about fixed-odds gambling. Looks like your formula's true by definition.
     
  15. kingwinner Registered Senior Member

    Messages:
    796
    9) A bag has 6 white balls and 4 red balls. Your randomly select 3 balls at the same time. What is the probability that all three balls will be white?

    My textbook do it this way:
    P(white)=n(white)/n(S)
    =(6C3 * 4C0) / 10C3
    =1/6

    But I wonder if that's true...my point is that there are identical objects (e.g. 6 white balls), would that affect the answer? How can I take care of the identical objects?
     
  16. przyk squishy Valued Senior Member

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    3,203
    The answer's correct (your probability or math textbook seems to be more reliable than your physics one). The identical objects are already taken care of within the formula. Its why the numerator has <sup>6</sup>C<sub>3</sub> * <sup>4</sup>C<sub>0</sub> instead of just 1.

    Another way of solving the problem is to consider the probability that the first ball you pick is white * the probability that the second is white given one white ball has already been removed * the probability the third is white given 2 white balls have already been removed = <sup>6</sup>/<sub>10</sub> * <sup>5</sup>/<sub>9</sub> * <sup>4</sup>/<sub>8</sub>. You get the same answer.
     
    Last edited: Apr 21, 2006
  17. kingwinner Registered Senior Member

    Messages:
    796
    The number of ways in which all three balls will be white should be only 1 possible way, right? WWW, if you interarrange the "W"s the result is the same because they are identical objects. However, 6C3 gives 20. How can that be right? I am so confused now

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  18. przyk squishy Valued Senior Member

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    If you number the white balls W<sub>1</sub>, W<sub>2</sub>, etc, you get 20 possible selections of 3 of these white balls. Since they're all indistinguishable, we have to add all of them.
     
  19. kingwinner Registered Senior Member

    Messages:
    796
    9) But in this case the balls having the same colour are indistinguishable. (you can't tell the difference between the 6 white balls)

    As a result,
    n(3 white), the number of possible ways obtaining 3 balls won't be = (6C3 * 4C0) or 20 and n(Universal set), all possible ways won't be = 10C3 or 120, right?

    If we consider the same-coloured balls as "identical" objects, how can I calculate the probability that all 3 balls will be white?



    First, let's consider a question like this:
    "A bag has 6 white balls and 4 red balls. Your randomly select 3 balls at the same time. How many possible combainations of 3 balls are there?"

    How can I calculate the combinations with identical objects? I don't remember how to do this...
     
  20. przyk squishy Valued Senior Member

    Messages:
    3,203
    Ultimately, four: 3 white balls, 2 white & 1 red, 1 white & 2 red, and 3 red balls.

    That doesn't mean the probability for each one is 1/4.

    If you number the balls 1 - 10, and the balls that interest you (ie. the white ones) are 1 - 6, the probability of selecting 3 of the first six balls is: the number of possible selections of 3 out of the first 6 (ie. <sup>6</sup>C<sub>3</sub>) divided by the total number of selections of 3 of the 10 balls (ie. <sup>10</sup>C<sub>3</sub>).

    The fact that there are 6 indistinguishable white balls means that more selections fit the criteria.
     
  21. kingwinner Registered Senior Member

    Messages:
    796
    "That doesn't mean the probability for each one is 1/4."
    I think I am understanding more, but my brain always tell me to be careful and take care of identical objects...

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    I have to think this stuff over again after posting this message...

    In my logical sense,
    Indistinguishable means W,W,W, you can't tell which is which
    Distinguishable means W1,W2,W3, i.e. you can tell which is which
    So the possible combinations for the case it contains indistinguishable (or identical) obejcts will be less.



    10) How many 3 letter subsets can be formed using the letters from the word "BOOKS"?
    Is there any formula to solve this? I don't recall any formula involving combinations with same objects identical. I only remember a formula for permutations with identical objects, that is n!/a!b!c!...
     
  22. kingwinner Registered Senior Member

    Messages:
    796
    This is another interesting way of solving this problem.
    Um...But the question says "Your randomly select 3 balls at the same time", not separately, will this method still work?
     
  23. przyk squishy Valued Senior Member

    Messages:
    3,203
    Basically, you have 10 balls (duh). Let's forget about colours for a minute and just label them 1, 2, 3, ... , 10. Let's say balls 1-6 are 'white.' Asking for the probability of picking 3 'white' balls is the same as asking for the probability of picking 3 of the first 6 balls. If balls 1-6 are indistinguishable in the context of the problem, it just means that you don't care whether you get balls 1,2,3, or 4,5,6, - so you just add up all the possibilities.

    Another way of phrasing the question is: "What's the probability that I'll select balls 1, 2, & 3, OR balls 4, 5, & 6, OR balls 2, 4, & 6, OR [17 other combinations]?"
    You've already solved problems like this - split up the combinations based on the number of O's they contain. Or if all else fails, apply Pete's algorithm

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