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Math Help - probability??

  1. #1
    Newbie ladiegimmick's Avatar
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    probability??

    I took a test at the begining of the school year, and my overall math score is 90%. I was recently looking at my grade book and I found out that I am struggling in a few areas. Can anyone show me how to complete these problems? You don't have to tell me the answers, I just want to learn how to complete them before next school year^_^

    1.) What is the experimental probability that exactly 3 children in a family of 4 children will be boys? Assume that P(boy) = P(girl).

    2.) The results of a coin toss are shown. What is P(heads)?
    HTHHHTHTTHHTHTT
    THHTHTTHHHHTHTT

    3.) A drawer contains 4 red socks, 3 white socks, and 3 blue socks. Without looking, you select a sock at random, replace it, and select a second sock at random. What is the probability that the first sock is blue and the second sock is red?

    4.) A lunch menu consists of 5 different sandwiches, 2 different soups, and 5 different drinks. How many choices are there for ordering a sandwich, a bowl of soup, and a drink?

    5.) The Burger Diner offers burgers with or without any or all of the following: catsup, lettuce, and mayonnaise. How many different burgers can you order?

    These aren't the exact questions on the test, I tried to make them as close as possible. This is were I am struggling. Permutations, combinations, and probability. :*(
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ladiegimmick View Post
    1.) What is the experimental probability that exactly 3 children in a family of 4 children will be boys? Assume that P(boy) = P(girl).
    We assume P(b)=P(g)=0.5.

    Then the number of boys in a family of 4 is a random variable with a binomial distribution with p=0.5 and N=4.

    So:

    P(3\text{\ boys})=b(3;4,0.5)= {4 \choose 3} 0.5^3 0.5=4\times 0.5^4=\frac{1}{4}

    Alternatively consider all the families with 4 children of whom 3 are boys in birth order:

    bbbg
    bbgb
    bgbb
    gbbb

    These are all equally likely, as are all families, so there are 4 families with 3 boys out of a total of 2^4 possible families, so teh required probability is:

     <br />
P(3 \text{\ boys})=\frac{4}{2^4}=\frac{1}{4}<br />
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