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Math Help - Help with some probability

  1. #1
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    Help with some probability

    a)Twelve people are selected randomly from a group of 50 to serve in a trial jury. Suppose Tim and Jane are two people in the pool of 50 people. What is the probability that both Tim and Jane will serve on the jury.


    b) A fair six- sided die is tossed 10 times and the resulting sequence is recorded. Let E be the event that the sequence of 10 outcomes contains exactly 4 ones, and let G be the event that the sequence of 10 outcomes contains exactly 6 threes. E and G can occur simultaneously.

    c) Cards numbered 1-13 are well shuffled and 5 cards are drawn and placed in a row from right to left. What is the probability that all 5 cards are odd numbered.

    d) A fair coin is tossed, a fair six-sided die is tossed, and three cards are drawn one at a time and without replacement from a standard 52 card poker deck. What is the probability that a 4 is tossed and three red cards are drawn.
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  2. #2
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    1.

    2. not sure what you are asking here, can you reiterate?

    3. There are a total of 7 odd numbered cards in the stack of 13, therefore the probability of you drawing the first odd number is \frac{7}{13} and the probability of you drawing a second odd is \frac{6}{12} and so on. so you would have:
     \frac{7}{13} \times \frac{6}{12} \times \frac{5}{11} \times \frac{4}{10} \times \frac{3}{9}= \frac{7}{429}

    4. Focusing on the die, the probability of getting a 4 using is \frac{1}{6}

    now for the cards, there are 13 red cards in the pile, so then drawing three we have:

    \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} = \frac{1716}{132600}

    since the events are independent we have:

     \left( \frac{1}{6} \right) \left( \frac{1716}{132600} \right) = \frac{11}{5100}
    Last edited by lllll; April 30th 2008 at 11:49 PM.
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