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Math Help - Probability Permutations and Combinations

  1. #1
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    Probability Permutations and Combinations

    Hello there,

    I have quite a few questions that I need some help for. I don't specifically require the answers, just the means/methods for figuring them out, then I can reply with my attempts. There's a few:

    1) A committee is selected from 7 men and 6 women.
    i) How many ways are there for a committee consisting of 3 men and 3 women?
    ii) How many ways for a committee of size 6 with one particular man and one particular women?

    2) i) In how many ways can the letters "abceeee" be arranged so that no "e" is adjacent to another "e"?
    ii) There are 20 seats on a bus. In how many ways can 3 people be seated in the bus?

    3) a) A box contains 10 memory sticks of which 4 are defective. Three memory sticks are drawn randomly without replacement. Find the probability that:
    i) they are all defective
    ii) only one is defective.
    iii) none of them are defective.
    (apparently supposed to be shown in real numbers to 3s.f and simplified rational numbers).

    4) A fair coin is tossed three times. Cnsider the events:

    A. TAILS on second toss
    B TAILS on third toss
    C TAILS on exactly one coin

    i) Find P(A), P(B) and P(C)

    ii) Find P(A n C) and decide whether it represents an independant event or not.





    Again, the methods or some guidance is good. I'd like to learn this myself. Thanks will be given!

    Regards,
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  2. #2
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    1) (i) Note that the order in which you select the members of the commitee does not matter, so you just need to calculate all the options of choosing 3 men out of 6 and multiply that by all options of choosing 3 women out of 7. This is (6C3)*(7C3) = 6!/3!3! * 7!/3!4! = 6!7!/3!3!3!4!

    (ii) First we'll need to pick one men and one women. In how many ways can we do this? We pick one man out of 6, so six ways for that, and for every choice of man we have 7 different women we can pick. So we have 6*7=42 ways of choosing one man and one woman. Now, you do the rest the same way as the earlier case. Can you solve it?

    2) (i) we can easily observe that the letters a,b,c will have to be spread between the e's, ie the word will be of the form: e_e_e_e. So we are only interested in how many ways we can arrange a,b,c in three slots. Can you follow from here?

    (ii) If you number the seats 1,2,...,20, then this problem is equivalent to simply choosing three numbers out of 20, with order being relevant, ie. for set of three sits that you will choose, there are 3! = 6 ways of arranging the students within these seats. There are 20C3 = 20!/17!3! ways of choosing 3 seats out of 20, and we will multiply that by 3! because the order matters. so 6*(20C3).

    3) (i) Four of the ten sticks are defective, so when we start from blank there is a 40% chance that we will pull a bad stick. All sticks being defective means we pulled out 3 defective sticks in 3 successive pulls. In the first pull, the chance of pulling a bad stick is 0.4 as mentioned. In the second, assuming we pulled a bad stick in the first time, there are 9 sticks left and 3 are defective, so 0.33 chance of pulling a bad one, hence there is a 0.4 * 0.33 chance of pulling two bad sticks in a row. Can you complete this up to three defective sticks?

    (ii) We want only one defective stick to be pulled out. The probability of taking out a bad stick the first time is 0.4, as mentioned. Now we want no bad stick the next times; the probability of this happening, in a similar fashion to the last excersize, is 0.66*0.625, so the probability of pulling one bad stick, where we pull it in the first time, is 0.4*0.66*0.625. Now calculate the probability of this happening in the two other cases and add them together to get the solution.

    (iii) We want none of them to be defective; this is solved in the exact same way as the first one. Can you do this one alone?

    4) (i) Getting tails on the second and third toss is irrelevant of the other tosses, since all that matters is that you get tail in this specific toss. Hence P(A) = P(B) = 0.5
    For P(C), we want a tails on exactly one coin toss. There are three different ways of this happening: tails 1st, heads 2nd and 3rd; heads 1st, tails 2nd, heads 3rd; heads 1st,2nd and tails 3rd. So we need to calculate the chance of each of these happening and add them.

    The probability of tails,heads,heads is 1/2*1/2*1/2 = 1/8. Can you do the rest from here?

    (ii) We want three tosses that will be: heads,tails,heads. The chance for each toss is 1/2, so we end up with 1/2^3 = 1/8. I don't know what an independant event is so I can't really answer that :X

    It's been three years since I've done any probability (and only high school level at that), so you may want to double-check my answers there.

    As for the basic counting problems, I find that simply using common sense is mostly the easiest way of solving the problem. You just need to know what you're counting, and make sure it makes sense.

    Hope this helped!
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  3. #3
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    Hey man, I'm done for the day today... huge headache. Will check this relatively early tomorrow so please refer back to it. Thanks so much for the help!
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  4. #4
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    Hey guys,

    I'm having some trouble with this... could anybody be more expansive in the explanations please? Rep/Thanks will be given for your time!

    Regards,
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  5. #5
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    Quote Originally Posted by rel85 View Post
    Hey guys,

    I'm having some trouble with this... could anybody be more expansive in the explanations please? Rep/Thanks will be given for your time!

    Regards,
    Please state what part post #2 you need clarification of.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Please state what part post #2 you need clarification of.
    Question 3ii) 4i and ii) please, with a relatively easy explanation.


    Also for for 3i) and 3iii) are the following answers correct respectively? 0.033 and 0.165.

    Regards,
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  7. #7
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    I assume you meant \frac{1}{30} and \frac{1}{6} respectively, if so, yes.

    I'll try to explain those questions better.

    3 (ii): There are three ways in which you can pull out only one bad stick, within three pulls:
    bad, good, good
    good, bad, good
    good, good, bad

    We want to calculate the probability of each of these happening, and add the three of them to get the total probability.
    The probability of the first happening, is \frac{4}{10} * \frac {6}{9} * \frac{5}{8} = \frac{1}{6} and so are the probabilities of the second and third, so we get that the total probability is \frac{1}{6} + \frac{1}{6} + \frac{1}{6} =\frac{1}{2}

    4 (i): A: We want a tails on the second toss. There are two possibilities of this happening:
    tails, tails
    heads, tails

    and there is a total of 4 possible distinct tosses, so P(A) = \frac{2}{4} = \frac{1}{2}

    B: We want tails on the third toss. There are 2^3 = 8 distinct toss "routes", out of these there are 4 possibilites of getting tails on the last toss:
    heads, heads, tails
    heads, tails, tails
    tails, tails, tails
    tails, heads, tails

    So P(B) = \frac{4}{8} = \frac{1}{2}

    C: There are three possibilities of tossing tails only once in three times:
    tails, heads, heads
    heads, tails, heads
    heads, heads, tails

    Out of 8 possible "routes", as I mentioned in B, so we have P(C) = \frac{3}{8}

    (ii): P(A \cap C) is the probability of both A and C happening at the same time; there is only one way this would happen:
    heads, tails, heads (we want exactly one toss of tails and we want it to be in the second toss)

    This is one of out 8 possible "routes", so P(A \cap C) = \frac{1}{8}
    Last edited by Defunkt; August 26th 2009 at 07:16 AM.
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