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Math Help - Probability & Expected Value

  1. #1
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    Probability & Expected Value

    I have a 4-question worksheet and I think I have them all correct except for one:

    1) A sorority sold 65 tickets in a raffle for a $480 TV set. What is the expected value of a single ticket if only one ticket wins?

    - I'm stumped.

    The other 4 questions I believe are correct--could someone confirm?

    2) A box contains 4 blue cards and 3 red cards. If 2 cards are drawn one at a time, find the probability that both are red with replacement and without replacement.

    - 9/49 with replacement
    - 1/7 without replacement

    3) Two dice are rolled. What are the odds against rolling a sum of 4 or 11?

    - 31/5

    4) In a NASA rocket firing, the probability of success of the 1st stage=70%, 2nd stage=56%, and 3rd stage=80%. What is the probability for success fir the three stage rocket?

    - (70/100) x (56/100) x (80/100) = 196/625

    Thank You!
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  2. #2
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    (1) there are 65 tickets in the raffle for the $480 tv. So the probability of winning is 1/65... so expected value of a ticket is E(V)=Pr(W)\cdot 480=\frac{1}{65} \cdot 480=\$7.38

    (2) Probability without replacement;
    \frac{2}{7}\cdot \frac{1}{6}=\frac{1}{21}
    Probability with replacement;
    \frac{2}{7}\cdot \frac{2}{7}=\frac{4}{49}

    3 and 4 look fine to me..
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  3. #3
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    Quote Originally Posted by mdenham2 View Post
    I have a 4-question worksheet and I think I have them all correct except for one:

    1) A sorority sold 65 tickets in a raffle for a $480 TV set. What is the expected value of a single ticket if only one ticket wins?

    - I'm stumped.

    The other 4 questions I believe are correct--could someone confirm?

    2) A box contains 4 blue cards and 3 red cards. If 2 cards are drawn one at a time, find the probability that both are red with replacement and without replacement.

    - 9/49 with replacement correct
    - 1/7 without replacement correct

    3) Two dice are rolled. What are the odds against rolling a sum of 4 or 11?

    - 31/5

    4) In a NASA rocket firing, the probability of success of the 1st stage=70%, 2nd stage=56%, and 3rd stage=80%. What is the probability for success fir the three stage rocket?

    - (70/100) x (56/100) x (80/100) = 196/625

    Thank You!
    \frac{\binom{3}{1}}{7}\frac{\binom{3}{1}}{7}=\frac  {9}{49}

    \frac{\binom{3}{2}}{\binom{7}{2}}=\frac{3}{21}=\fr  ac{1}{7}
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