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    Multiple questions in probability and statistics

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    Can you please provide step by step instruction on how to work these problems? I have tried to worked them out, but do not get the answers posted. Attached are the problems. Thanks!

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    See the attached file.

    27. Suppose 2 cards are drawn without replacement from an ordinary deck of 52. Find the probabilities of the following results. (Answer: .9955)

    At most one is a queen.

    Note that first card can be drawn in 52 ways and second in 51 ways. So total ways = 52 x 51

    Since there are 4 queens in a pack, First queen can be drawn in 4 ways and the next in 3 ways. So total ways to pick 2 queens = 4 x 3 = 12.

    The chance that both cards are queens = (4 x 3)/ (52 x 51) = 0.00452488

    Chance of at most 1 queen = 1 - 0.00452488 = 0.9955 nearly

    47. In placebo-controlled trials of Prozac®, a drug that is prescribed to fight depression, 23% of the patients who were taking the drug experienced nausea, whereas 10% of the patients who were taking the placebo experienced nausea.*

    a. If 50 patients who are taking Prozac® are selected, what is the probability that 10 or more will experience nausea?

    b. Of the 50 patients in part a, what is the expected number of patients who will experience nausea?

    c. If a second group of 50 patients receives a placebo, what is the probability that 10 or fewer will experience nausea?

    d. If a patient from a study of 1000 people, who are equally divided into two groups (those taking a placebo and
    those taking Prozac), is experiencing nausea, what is the probability that he/she is taking Prozac?

    e. Since .23 is more than twice as large as .10, do you think that people who take Prozac® are more likely to experience nausea than those who take a placebo? Explain.

    (Answers: a. about .74 b. about 12 c. about .99 d. about .70)

    The distribution in this case is a binomial distribution.
    The binomial distribution consists of n trials and p, the probability of success in each trial. Chance of exactly

    k successes = (n choose k) p^[k](1 - p)^[n-k]

    Where ^ means raised to the power of.

    (n choose k) = n! / (k! (n - k)!) where ! means factorial.

    a) Probability of nausea = 0.23
    Chance that exactly k patients experience nausea is given by:
    (50 choose k) x 0.23^k x (1 - 0.23) ^ (50 - k) where ^ means raised to the power of.

    Since we want the probability for k >= 10, we sum ...

    Solution Summary

    Probability of card drawing
    Binomial distribution
    Probability in Lottery