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Guys please help with Statistics.

  1. Jul 23, 2006 #1
    this is for extra credit in this summer class that i am taking at WVU. The teacher said he doubts we get them all right but to look for help if you can. Ive already posted this at another forum and have some answers already....if you can help or add to the problems please do!

    guys here is the whole worksheet if you want to look over it......its in condensed form.

    1)In Yahtzee, you get 3 rolls of the dice per turn, and you get to choose which ones you re-roll each time. My brother calls me up and asks me the probability of him winnin gthe game. If he gets a large straight, he wins, otherwise my Dad wins. With his first roll, he got 2,3,4,6,6. He is going to re-roll the second 6 that he got, trying to get a 5. He has two rolls to get a 5. What's the probability that he gets that 5 and beats my Dad.
    "For question 1:

    1/6 + (5/6 x 1/6)
    = 1/6 + 5/36
    = 11/36"

    2) You and i are going to play a game. I am going to be flippin a coin, and you will be rolling a die. i get to go first. If i get heads, i win the game; whereas if you roll the die and get the number 1,2,3 or 4 you win. It goes back and forth until someone wins. Who has the better chance of winning, and why?
    "For number 2 the answer is pretty obvious.

    Whoever rolls the die has the better chance.
    They have 50% to win right off the bat.
    Whereas the person who rolls the die has to not only win the toss, but he also has to roll the right number on his turn."

    3) Almost the same game as above. This time I win if I get a 1,2,3,4, or 5. Who has the better chance and why?

    Im not sure how you all worked the math on number 2 so i dont know about number 3...i have a feeling he still has a better chance?

    4) My firend came down to visit me last semester, and wanted me to take them to the wVU vs. Virginia tech football game. Tickets were sold out, but i planned on asking my students' in Lab throughout the day to see if they had an extra ticket i could buy. I had 3 firends coming down, so i needed 4 tickets. If i had 125 students to ask throughout the day, and each one had a prob. of 5% chance of having an extra ticket, what is that probability that my friends' and I will get to go to the game?

    "For 4.
    You need at least 4 tickets.
    So do 1-P(not getting 4 tickets)
    So 1 - p(3) - p(2) - p(1) - p(0)
    The probability of getting 3 is :
    (0.05)^3 x (0.95)^122 x 125C3
    And you repeat for 2, 1, and 0."

    5) After a hard fough fooball game, it was reported that, of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and arm, 2 hurt both a hip an dknee, 1 hurt a arm and knee, and no one hurt all three. Comment on the accuracy of the report.
    not possible

    6) There are 5 red chips and 3 blue chips in a bowl. The red chips are numbered 1,2,3,4,5 respectively. The blue chips are numbered 1,2,3, repectively. If 2 chips are drawn at random and without replacement, whats the prob. that these chips have either the same number or the same color?

    ".5712 is the same as 57.14 so i guess im good there."

    7) Machines I, II, and III produce springs. Machine I has a 1% chance of defect, Machine II has a 4% chance, and Machine 3 has a 2% chance. Machine I produces 30% of the springs, Machine II produces 25%, and Machine III produces 45%. Whats the prob. of randomly selecting a defective spring.

    For number seven i got .022 or 2.2 percent...but take it for what its worth.

    8) Players A and B play a sequence of independent games. Player A throws a die first and wins on a "six". If he fials, player B throws and wins on a "five" or "six". If he fails, A throws and wins on a "four" "five" or "six". And so on... Find the proability of each player winning.

    "For question 8:
    A has (1/6) + (5/6 x 4/6 x 3/6) + (5/6 x 4/6 x 3/6 x 2/6 x 5/6)
    B has (5/6 x 2/6) + (5/6 x 4/6 x 3/6 x 4/6) + (5/6 x 4/6 x 3/6 x 2/6 x1/6)
    I think that's right.
    And you just do the math from there.
    52.15% for A
    47.77% for B
    Doesn't seem right to me.. But meh"

    9) The five of us are going to play Yahtzee. What's the prob. that I win?
    1/5 ?
  2. jcsd
  3. Jul 24, 2006 #2
    This answer doesn't make much sense. I assume you mean to say that the person who flips the coin has the better chance of winning, but the answer doesn't look deeply enough to say that for sure (though in this case the answer is correct).

    First consider the player who is flipping the coin. On his first flip, his chance of winning is .5. On his second flip, he needs to have missed his first flip (.5 probability), and he needs the other player to have rolled a 5 or a 6 (1/3 probability). You can keep going like that, and you'll find that the probability that the flipping player wins is

    [tex]P = \sum_{i=1}^\infty \left(\frac{1}{2}\right)^i \left(\frac{1}{3}\right)^{i-1}.[/tex]

    That's just a geometric series which you can easily evaluate to get [itex]P=0.6[/itex]. The probability that the rolling player wins is then just [itex]1-P=0.4[/itex] (since one or the other wins). So here the flipping player will probably win.

    Question 3 is similar. According to my arithmetic, your intuition is correct (but I'm not very good at arithmetic, so check yourself!!!). Edit: Sorry, your intuition is obviously correct. I'm silly!

    Your answer here is correct, but can you explain why?

    The rest of your answers also look fine to me.

    EDIT: woops, forgot to look at #6. That one looks pretty sketchy!

    EDIT #2: ok, #6 looks alright too (except you probably want to explain why that's the answer)!
    Last edited: Jul 24, 2006
  4. Jul 24, 2006 #3


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    #2 is basically fine the way you answered (assuming you meant "the coin has a greater chance to win"). You don't need to calculate. All you have to know is:
    1. The coin goes first, giving it an instant 50% chance to win
    2. Even if it misses the first flip, it has a greater than 0 chance to win on some other flip
    3. Therefore its total chance to win is greater than 50%.
    This is basically what you said and it also goes for question #3 assuming the coin goes first.

    #6 doesn't look right to me. I think the answer is .464
    The rest are good, but yes you should explain your reasons for #5.
  5. Jul 24, 2006 #4
    as for 6, i think:

    look for the sign OR --> it means addition of 2 separate chances.

    chance for 2 the same colors: (5/8 * 4/7) + (3/8 * 2/7) = 13/28

    chance for two the same numbers: possible combinations: 11/22/33

    --> chance of 1 such combination = 1/8 * 1/7 = 1/56, thus total chance is 3/56

    --> total chance = 13/28 + 3/56 = 0,518
  6. Jul 24, 2006 #5


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    :grumpy: I forgot to include the red-red case.
  7. Jul 24, 2006 #6
    this extra credit is due this morning so you guys probly wont have time to respond but....

    number 6 is done like this.

    same color (3/8)(2/7)
    same color (5/8)(4/7)
    same number ((2/8)(1/7)) x 3 (2/8 is for one pair of the three possible pairs... i got my answer then multiplied by 3 to account for every pair.)


    .1071 + .35714 + .1071 = .57134
  8. Jul 24, 2006 #7
    ah, yes! I'm silly :frown:

    I think you missed the "or the same number" piece :smile:
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