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Homework Help: Please help with probability and statistics problem!

  1. Aug 14, 2012 #1
    These questions are from a textbook. I attempted them, but couldn't get the answer. Please try which ever ones you can and provide a brief explanation if necessary of how to get the answer.

    1. What are the odds against your rolling doubles in fewer than four rolls of dice?

    First, I tried to find the probability of not rolling doubles which is 5/6.
    So, for not rolling doubles, it'd be (5/6)^4. I'm not sure what to do...

    Ans: 125:91

    2. What is the expected number of people you would have to sample to find someone with the same birthday as yours?

    p = 1/365
    q = 364/365
    Expected number of people to ask before finding the same birthday: q/p = 364
    Add one to find someone with the same birthday???

    Ans: 364

    3. If a recent poll indicated that the Hippopotamus party had 9% popular support, what would be the expected number of people that a television interviewer must survey to obtain the opinion of a Hippo supporter?

    p = .09
    q = 0.91
    E(X) = q/p = 10.111
    Round up to 11?

    Ans: 10

    5. A child removes the name cards on ten wrapped Christmas gifts of which six are for the Changs and four are for the Andersens. If the presents are distributed to the two families without reopening the packages,
    a) what is the probability that the Changs get five of the presents originally intended for them? Ans: 12/105
    b) What is the expected number of correct presents distributed to the Changs? Ans: 3.6

    6. Lesley claims she can roll triples with three dice in fewer than 20 rolls?
    a) What is the expected number of rolls that it would take to roll the first set of triples?
    E(X) = q/p =35 (before you get first triple, so wouldn't you add one?)

    Ans: 35
    b) What is the probability that she will roll the triples in fewer than five rolls?
    Ans: 0.107
  2. jcsd
  3. Aug 14, 2012 #2

    Ray Vickson

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    Homework Helper

    1. Presumably, fewer than 4 means at most 3, so P{no doubles in 3 tosses} = (5/6)^3. Do you know how to get P{at least one double in 3 tosses} from P{no doubles in 3 tosses}? (It is, of course, possible that the author of the book was careless and really meant 4 or less, rather than < 4; however, I would hope this was not the case.)

    2. Assuming uniform birthdates across a year, p = P{same b'day as you} = 1/365, as you said, and of course, q = 1-p = 364/365. You wrote down an answer with no explanation, and I am concerned that you may not (or you may) understand exactly what you should do. So, please supply more details; for example, is the number you need to sample a random variable with a familiar, recognizable distribution?

    3. Again, what distribution are you using? Are you sure it is EX = q/p, rather than EX = 1/p? And no: do not round up or down: the expected value has a well-defined technical meaning, and so you are not free to alter it to your taste. EX need not be a whole number.

    4. ---missing!

    5. Do you recognize the distribution used here? You have a population of 10 items, 4 of type A and 6 of type C. You deal 4 at random (to the Andersons---the Changs get the rest). You want P{deal 4 of type A}. This has a familiar distribution; do you recognize it? (a) When you say ans: 12/105, does that mean _your_ answer, or the answer at the back of the book? If it is yours, show your work. (b) Show the work.

    6. This is not a question, it is an assertion, so don't use "?". Anyway, show how to get p = P{at least one triple in one toss of 3 dice}; you said p = 1/36, but you should demonstrate how you got it. Again: the old question---should you use q/p or 1/p? You need to understand exactly what random variable X has EX = q/p and what random variable Y has EY = 1/p; then you need to understand whether to use X or Y. If you understand exactly what X and Y represent, there will be no problem making the correct decision.

  4. Aug 14, 2012 #3
    The "Ans:" are the answers from the book.

    1. Yes, I agree with you that it's under four rolls, so it'd be (5/6)^3 = 125/216. 216-125 = 91, but how do I relate these two to get odds?

    2. This question is involving waiting times with a geometric distribution. That is P(X=k)=(q^k)(p). Where X is the number of trials before a success.

    3. Same as question two.

    4. Sorry about that.

    5a) I was thinking, since there are six gifts for the Changs, it'd be like 6C5 (6 choose 5), divided by 10C5 for the total ways the gifts can be distributed, but this seems to be the wrong answer.
    5b) I'm not sure how to do this!

    6b) Would this be a binomial distribution?
  5. Aug 14, 2012 #4

    Ray Vickson

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    1. Google is your friend; see, eg.,

    2. I think you are using the wrong distribution. The _usual_ definition of the geometric distribution is that it counts the number of trials until (that is, including) the first success, so is P{X = k} = p q^(k-1), k=1,2,..., with EX = 1/p. The one you are referring to is another form of "geometric" distribution, but is only rarely used. It, of course, counts the number of trials _before_ the first success, and has distribution P{X=k} = p*q^k, k=0,1,2, ..., with EX = q/p. My reading of the question is that the success trial is also counted.

    3. Same as Q2.

    5. (a) If the Changs get presents CCCCCA in that order, the probability is (6/10)(5/9)(4/8)(3/7)(2/6)(4/5), because the first has prob 6/10, leaving 9, of which 5 are for C, etc. However, the order could also be ACCCCC or CACCCC, etc., so the probability is 6 times the above, giving p = 4/35 = 12/105.

    (b) The short answer is NO: it is not a binomial distribution (it is a *hypergeometric distribution*); however, the expected value is the same as if it were a binomial! See, eg.,
    http://mathworld.wolfram.com/HypergeometricDistribution.html for some basic material about the hypergeometric distribution.

    Last edited: Aug 14, 2012
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