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Math Help - Odds

  1. #1
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    Exclamation Odds

    Aki is a participant
    on a trivia-based game show. He has an
    equal likelihood on any given trial of being
    asked a question from one of six categories:
    Hollywood, Strange Places, Number Fun,
    Who?, Having a Ball, and Write On! Aki
    feels that he has a 50/50 chance of getting
    Having a Ball or Strange Places questions
    correct, but thinks he has a 90% probability
    of getting any of the other questions right. If
    Aki has to get two of three questions
    correct, what are his odds of winning?
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  2. #2
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    I tried finding all the different combinations and using the probabilities but cannot get the right answer
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  3. #3
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    Quote Originally Posted by ibetan View Post
    Aki is a participant
    on a trivia-based game show. He has an
    equal likelihood on any given trial of being
    asked a question from one of six categories:
    Hollywood, Strange Places, Number Fun,
    Who?, Having a Ball, and Write On! Aki
    feels that he has a 50/50 chance of getting
    Having a Ball or Strange Places questions
    correct, but thinks he has a 90% probability
    of getting any of the other questions right. If
    Aki has to get two of three questions
    correct, what are his odds of winning?
    First, if he has an equal likelihood of being asked a question from any of the 6 categories,
    and he has to answer at least 2 of 3 questions correctly,
    the situation can be divided up according to the questions he must answer.

    3 "Having a Ball" questions.
    2 "Having a Ball" questions and 1 "Strange Places" question.
    1 "Having a Ball" question" and 2 "Strange Places" questions.
    3 "Strange Places" questions.
    1 "Having a Ball" question, 1 "Strange Places" question and 1 "other four" question.
    2 "Having a Ball" questions and 1 "other four" question.
    1 "Having a Ball" question and 2 "other four" questions.
    2 "Strange Places" questions and 1 "other four" question.
    1 "Strange Places" and 2 "other four" questions.
    3 "other four" questions.

    then, beginning with the first category (3 Having a Ball questions), you can calculate the probability that
    he is asked the question and he answers 2 correctly or 3 correctly.
    If he answers 2 correctly, then he can get the first, second or third one wrong.

    Hence that probability he'd get 2 correct would be

    \frac{1}{6}\ \color{red}0.5\color{black}\ \frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5+\frac{1}{6}\ 0.5\ \frac{1}{6}\ \color{red}0.5\color{black}\ \frac{1}{6}\ 0.5+\frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5\ \frac{1}{6}\ \color{red}0.5

    =\left(\frac{1}{6}\right)^3(0.5)^2\color{red}(0.5)  \color{black}(3)=\left(\frac{1}{6}\right)^3(0.5)^3  (3)

    and the probability of getting all 3 correct is

    \frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5=\left(\frac{1}{6}\right)^3(0.5)^3

    Continuing on with the other categories, the remaining probabilities can be calculated and then summed.
    Last edited by Archie Meade; March 19th 2010 at 04:50 AM.
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  4. #4
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    Hello ibetan
    Quote Originally Posted by ibetan View Post
    Aki is a participant
    on a trivia-based game show. He has an
    equal likelihood on any given trial of being
    asked a question from one of six categories:
    Hollywood, Strange Places, Number Fun,
    Who?, Having a Ball, and Write On! Aki
    feels that he has a 50/50 chance of getting
    Having a Ball or Strange Places questions
    correct, but thinks he has a 90% probability
    of getting any of the other questions right. If
    Aki has to get two of three questions
    correct, what are his odds of winning?
    If we group the questions into 'hard' and 'easy', then there are 2 categories in the 'hard' group (Having a Ball and Strange Places) and 4 categories in the 'easy' group (the remainder of the 6 categories).

    Assuming that the categories are chosen at random, the probability that a 'hard' question is chosen is therefore \tfrac13, and the probability that an 'easy' question is chosen is \tfrac23.


    If a 'hard' question is chosen, the probability that Aki gets it right is \tfrac12; and if an 'easy' question is chosen, the probability he gets it right is \tfrac{9}{10}.


    So the probability that Aki gets a question chosen at random right is:
    \frac13\times\frac12 + \frac23\times\frac{9}{10}
    = 0.7667
    Another assumption now: Aki is asked all three questions, even if he gets the first two right. (The working will be a bit different if a third question isn't asked if this happens.)

    So the probability that he wins is the probability that, out of three questions, he gets all three right, or two out of three right
    =(0.7667)^3 + 3(0.7667)^2(1-0.7667)^1

    = 0.862 (to 2 d.p.)
    Grandad
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello ibetanIf we group the questions into 'hard' and 'easy', then there are 2 categories in the 'hard' group (Having a Ball and Strange Places) and 4 categories in the 'easy' group (the remainder of the 6 categories).

    Assuming that the categories are chosen at random, the probability that a 'hard' question is chosen is therefore \tfrac13, and the probability that an 'easy' question is chosen is \tfrac23.


    If a 'hard' question is chosen, the probability that Aki gets it right is \tfrac12; and if an 'easy' question is chosen, the probability he gets it right is \tfrac{9}{10}.


    So the probability that Aki gets a question chosen at random right is:
    \frac13\times\frac12 + \frac23\times\frac{9}{10}
    = 0.7667
    Another assumption now: Aki is asked all three questions, even if he gets the first two right. (The working will be a bit different if a third question isn't asked if this happens.)

    So the probability that he wins is the probability that, out of three questions, he gets all three right, or two out of three right
    =(0.7667)^3 + 3(0.7667)^2(1-0.7667)^1

    = 0.862 (to 2 d.p.)
    Grandad
    Thanks. I understand everything you did except the part in  3(0.7667)^2 , why did you multiply by 3?
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  6. #6
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    Hello ibetan
    Quote Originally Posted by ibetan View Post
    Thanks. I understand everything you did except the part in  3(0.7667)^2 , why did you multiply by 3?
    This is an example of the Binomial Distribution Formula. The probability of getting r successes from n trials is:
    \binom nrp^r(1-p)^{n-r}
    Here, p = 0.7667, \;n = 3,\; r = 2. The probability of getting 2 questions right out of 3 is:
    \binom 32(0.7667)^2(1-0.7667)^1
    and \binom32 = 3. This is where the 3 comes from.

    It is simply the number of ways of choosing 2 questions out of 3, these being the questions that Aki gets right. He could get:
    right, right, wrong
    or
    right, wrong, right
    or
    wrong, right, right
    3 ways. Do you see now?

    Grandad
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