More probability questions

Discussion in 'Physics & Math' started by kingwinner, Apr 29, 2006.

  1. kingwinner Registered Senior Member

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    796
    Here are some probability questions. I checked my work over and over again but I still can't find out where I went wrong. I can't get the answer. Can someone please help me? Thank you very much!

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    1) In a family of 4 children, determine the probabilities that the first is a girl OR the next 3 are boys.

    Let B be the event the first is a girl
    Let C be the event the next 3 are boys
    P(B U C)=P(B)+P(C)-P(B intersect C)
    =1/2 + (1/2)^3 - 1/16
    =9/16 or 0.5625<--but the answer provided is 0.6875, why?

    2) Of 100 circuit boards in a production run at a factory, 8 are defective. If a random sample of 5 is taken, what is the probability that at least 1 is defective?

    Let A be the event at least 1 is defective
    n(A')=92C5 * 8C0
    n(S)=100C5
    P(A')=n(A')/n(S)=0.653191
    P(A)=1-P(A')=0.3468 <---this is what I get but the answer provided is 0.3409. Where did I went wrong or is the answer provided wrong?

    3) On a Las Vegas roulette wheel, there are 18 black, 18 red and 2 green numbers. If it is spun twice, what is the probability that a green number will come up first OR a red number will come up second?

    n(S)=38^2=1444
    n(G)=2x37=74
    n(R)=37x18=666
    n(G intersect R)=2x18=36
    P(G U R)=P(G)+P(R)-P(G intersect R)
    =74/1444 + 666/1444 - 36/1444
    =704/1444
    =176/361 or 0.4875 <--but the answer provided is 0.5014, where did I went wrong this time?

    4) What is the probability that in a group of 10 students, at least 2 will have the same birth month?
    [The thing that gets me stuck is the fact that each month has a different number of days, e.g. 28, 30 ,31. Should I and can I ignore this fact and assume each month has the same number of days?]
     
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  3. przyk squishy Valued Senior Member

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    Hi kingwinner,

    The book mistakenly added the intersection instead of subtracting it. In general, if your answer is different from the book's, its a good idea to try some modifications to your calculation to see if you can guess how the book got its answer - it'll help you decide whether you or the book is correct.
    The correct answer depends on what is meant by 8 of 100 circuit boards being defective. It could mean that they have exactly 100 circuit boards, of which exactly 8 are known to be defective, or it could mean that on average 8 in every 100 produced are defective. Most likely they meant the latter, seeing as the answer for this is the one provided in the book.
    I highlighted the errors - should be 38 instead of 37. I don't see why you did such a complicated calculation, since P(G) is simply 2/38 and P(R) is 18/38.
    I'd assume the year is equally divided into 12 months, otherwise the calculation will become a nightmare.
     
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  5. kingwinner Registered Senior Member

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    796
    hi przyk,

    2) I guess they meant that exactly 8 out of the 100 are defective.
    But would the answer still be the same whether 8 is an average or not? I don't see how they get an answer of 0.3409

    3) Thanks for pointing out my error. I forgot the fact that they are independent events.
    "I don't see why you did such a complicated calculation, since P(G) is simply 2/38 and P(R) is 18/38." <----Why is P(G)=2/38? Don't you have to consider all the possibilities? For example, GBB is a different outcome than GBG, GRB, etc. so I calculated all the possible outcomes...
     
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  7. przyk squishy Valued Senior Member

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    It does make a difference whether the figure is an average or not. You assumed it wasn't and got 0.3468, the book assumed it was and got 0.3409 (which is correct - I checked). If the figure is an average it simply means that any untested circuit board has a 8/100 chance of being defective, regardless of how many defective boards have already been found. The calculation is actually simpler for this than if you assume it isn't an average. The problem is the question can be interpreted both ways, though.
    Do you think its a coincidence that <sup>76</sup>/<sub>1444</sub> = <sup>2</sup>/<sub>38</sub>, and <sup>684</sup>/<sub>1444</sub> = <sup>18</sup>/<sub>38</sub> ?
     
  8. kingwinner Registered Senior Member

    Messages:
    796
    2) If the question doesn't say average, I would interpret it as exactly 8 out of 100 are defective.
    But if it's average, how would you calculate the probability?


    3) No, I don't think so. But why would they be the same?
     
  9. przyk squishy Valued Senior Member

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    3,203
    If a company claims that 8 in 100 of their circuit boards are defective, they typically mean that every circuit board they manufacture has an 8% chance of being defective. So, if you pick 5 such circuit boards at random, what's the probability that you'll get at least one defective one?
    Look at your calculations: You got P(G) by calculating 2*38 / 38<sup>2</sup>. This is what you'd expect, since the 2 spins are independent. The second spin can't in any way affect the probability that the first spin gives a green.
     
  10. kingwinner Registered Senior Member

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    796
    2) But I think they may have taken the next two digits by accident (the probability is 0.346809...) which should be rounded to 0.3468 not 0.3409
     
  11. kingwinner Registered Senior Member

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    796
    I have some more questions!

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    5) A critical circuit in a communication network relies on a set of 8 identical relays. If any one of the relays fails, it will disrupt the entire network. The design engineer must ensure a 90% probability taht the network will nto fail over a 5-year period. What is the maximum tolerable probability of failure for each relay?

    I have seen 2 methods of solving this problem, but the answers aren't matching exactly which suggest that one of them may be wrong. Which of the following is the correct method?

    Method 1:
    Let x represent the probability that the individual relays don't fail
    Then 0.90 = x^8
    x=0.987
    Thus, the answer is 1-0.987=0.013 or 1.3%

    Method2:
    Let y represent the probability of any one of the relays failing
    Then 8x = 0.10
    x=0.0125 or 1.25%

    6) Of the members of a track-and-field club, 42% entered track events at the most recent provincial meet, 32% entered field events, and 20% entered both track and field events. What is the probability that a randomly selected member of te club entered either a track event or a field event at the provincial meet?

    P(T U F)=0.42 + 0.32 - 0.20
    = 0.54 <-however, the answer provided is 0.46, why?

    7) Naomi has 12CDs: 3 of the band Nine Inch Nails, 4 of the band Soundgarden, 2 of the band Monster Magnet, 1 of the band Pretty & Twisted, and 2 of the band Queensryche. If he randomly loads his player with 5 CDs, what is the probability that it will hold 3 Nine Inch Nails CDs OR 3 Soundgarden CDs?

    [(3C3 * 9C2) / 12C5] + [(4C3 * 8C2) / 12C5]
    =37/198 <-but the answer in my textbook is 13/66, why?
     
    Last edited: Apr 30, 2006
  12. przyk squishy Valued Senior Member

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    3,203
    I doubt it, but thats an amazing coincidence. Its pointless argueing though, since the original question was a little ambiguous.

    #5: Method 1 is correct. Where did you get the other one from?
    #6: I have no idea why they give the complement of your answer.
    #7: They added the possibility of 4 Soundgarden CDs being in the CD player.
     
  13. kingwinner Registered Senior Member

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    796
    5) I now understand that method 2 doesn't work because more than one relay may fail and that method doesn't include it.

    6) 7) So the answers from my textbook are both wrong...again!

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    Last edited: May 1, 2006
  14. kingwinner Registered Senior Member

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    796
    These 2 problems I have no idea how to do them. Can someone give me some hints?

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    8) 2 brands of headache remedy are on the market: Acetylin and Salicin. One in 400 people taking Acetylin suffers side-effects and one in 1200 taking Salicin. At the present time it is estimated that equal numbers of people take each kind of drug. If Acetylin is taken off the market because of industrial sabotage during the processing of the drug, show that the probability of side-effects will be halved.

    9) In a local school, 48% of the students are male and 52% are female. Portable radios are used by 15% of the males and by 24% of the females. A portable radio was turned into the lost and found. What is the probability that it is owned by a female student?


    Thank you!
     
  15. przyk squishy Valued Senior Member

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    3,203
    Even then the calculation's wrong. The probability of one of the relays failing is 8x(1-x)<sup>7</sup>.
    6 looks wrong, 7 looks ambiguous. They did the same calculation as you, and then added an extra term. I think its fairly safe to assume they did it deliberately. Anyway, if you have 4 Soundgarten (never heard of this band, or any of the others while we're at it) CDs in the CD player, there are still 3 Sg CDs in there, aren't there? You could still pull out 3 Sg CDs anytime you wanted.

    #8 You're being asked to compare the probability of a random person having side effects if 50% of people take aspirin and 50% take salicin, with the probability of a random person having side effects if everyone takes salicin.

    #9 I suggest you draw a tree diagram. It might help.
     
  16. kingwinner Registered Senior Member

    Messages:
    796
    5) I think for method 2 they did it this way:
    Probability of the 1st one failing + probability of 2nd one failing +... probability of 8th one failing
    Then x+x+...+= 8x = 0.10
    x=0.0125 or 1.25%
    And I think this is wrong because they did not include the probability of 2 of them failng the same time, 3 of them failing the same time, etc.

    7) But the question says "...what is the probability that it will hold 3 Nine Inch Nails CDs OR 3 Soundgarden CDs", so I would interpret it as exactly 3 Soundgarden CDs, make sense?
     
  17. kingwinner Registered Senior Member

    Messages:
    796
    10) A card is drawn form a deck 5 times and replaced each time. Determine the probability that the first 2 cards are king OR all 5 are red.

    I am not too sure what would the intersect of the 2 events be. I am confused.
    Is the intersection the first 2 cards are kings AND all 5 are red? (see method 1)
    Or is the intersection the first 2 cards are red kings? (see method 2)

    Let A be event first 2 cards are kings
    Let B be event all 5 cards are red

    Method 1:
    P(A U B) = P(A) x P(B) - P(A intersect B)
    =(4/52)^2 + (1/2)^5 - (4/52)^2 * (1/2)^5
    =0.03698

    Method 2:
    P(A U B) = P(A) x P(B) - P(A intersect B)
    =(4/52)^2 + (1/2)^5 - (2/52)^2
    =0.03569

    Which one is correct? I did it the method 1 way, but my friend did it the method 2 way, and both seem to make sense. But surely, only 1 is correct and I don't know which one.
     
  18. przyk squishy Valued Senior Member

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    3,203
    There are 2 ways of doing it the second way, if you want to do it correctly and get the same answer as 1:

    Method 2a: 8*P(one fails and the 7 others do no not fail) + <sup>8</sup>C<sub>2</sub>*P(2 fail and the 6 others do not) + [all the other possibilities].

    OR

    Method 2b: P(first one fails) + P(first one does not fail and the second one does) + P(1 and 2 do not fail and 3 does fail) + ...
    As I said, ambiguous question. Mathematics is precise, English is not.

    For that other problem:
    #10 The intersect is the probability that you get both conditions, ie. the first 2 are kings, and all 5 are red. Your method (#1) is correct

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    :bugeye: Don't you have a teacher for this kind of thing? You're posting a LOT of your homework on the internet...
     
  19. kingwinner Registered Senior Member

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    796
    Yes, of course I have a teacher. But it is not very possible to ask that many questions during the short class time because there is a new lesson every day. And I always do all my homework to make sure I am on top of this stuff. The questions I posted are only a small fraction of all my homework.
    Thank you for helping, by the way, if you don't mind(I hope).

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    Last edited: May 2, 2006
  20. kingwinner Registered Senior Member

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    796
    These 2 questions looks easy to me at first glance, but it seems to be harder than I was thought.

    11) Two cards are drawn from a standard deck. After the first card is drawn, it is not replaced before the second card is drawn. What is the probability that the first card is a queen and the second card is not a spade?

    My first attempt is (4/52) x (39/51)=1/17
    But one thing that troubles me right away is the fact that if the first card is a queen that is not a spade, the probability of the next card being not a spade will be changed, is that true?
    So the answer may be (1/52)(39/51) + (3/52)(38/51)=3/52, I think...

    12) There are 3 white balls, 4 green balls, 5 red balls, and 6 blue balls in a bag. Find P(red and blue) if 2 balls are drawn, with replacement being made after the first ball is drawn.

    (5/18)(4/18) <--do I have to multiply this by 2? (because I was thinking that of (red,blue) and (blue,red) I am not too sure...

    Am I on the right track?
     
    Last edited: May 2, 2006
  21. przyk squishy Valued Senior Member

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    3,203
    I see.
    No need to tell you that's an excellent idea of course...
    I don't, and no problem by the way

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    As for those new problems you posted:

    #11 You're on the right track with your second equation, but are you sure its (1/52)*(39/51)+... ?

    #12 Another sucky ambiguous question. As you noticed, there's a difference between drawing 1 red and 1 blue (in any order), and the first one being red and the second blue. I'm not really sure which one they meant - the answer book will tell you.
     
  22. kingwinner Registered Senior Member

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    796
    11) I kind of believe the answer is (1/52)(39/51) + (3/52)(38/51)=3/52 because the cards are drawn without replacement. Case 1 is when a spade queen is drawn, the probability of the next card being not spade will be 39/51. Case 2 is when a queen that is not a spade is drawn, so there will be 38 "not spade" cards left. I just added Case 1 and Case 2 to get the answer.
     
  23. przyk squishy Valued Senior Member

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    3,203
    Never mind - I was thinking "second card is a spade" instead of "second card isn't a spade, so I thought you'd put the <sup>1</sup>/<sub>52</sub> where the <sup>3</sup>/<sub>52</sub> should have been. My mistake :m:
     

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