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Math Help - Probability Help

  1. #1
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    Probability Help

    Hey guys I hope I enjoy my time here, just thought I would like to say that before I start this thread.

    I have 2 problems which I am having difficulty, the first one being:

    Problem 1:
    From a box containing six red balls and two white balls, one ball is drawn at random and then replaced.
    a) What is the probability that the ball drawn is white?
    b) If the experiment is performed three times, what is the probability that a white ball is drawn at least twice?

    Okay I know that a) is 2/8 which goes to 1/4. That was easy, but I'm not sure how to do B. I know the answer for B is 5/32 but I'm not sure how to get that.

    Problem 2:
    Gaylee has found that on a fine day she makes a sale to 70% of her client while on a rainy day she makes a sale to only 30% of her clients. Gaylee works 24 days in June and 9 of them are rainy.
    i) Draw a tree... I've done that
    ii) Calculate the probability that Gaylee makes a sale on any given working day in June. Done that, answer is 11/20
    iii) Gaylee has also found that in June she is four times as likely to make a sale when she wears her floral hat than when she doesn't wear a hat at all. Given that Gayle wears her floral hat on 16 of the 24 working days in June, what is the probability that she makes a sale given that shie is wearing her floral hat. I need help with this one. The answer was given to me, 11/15, but I am unsure in how to calculate it.

    Thanks
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  2. #2
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    Re: Probability Help

    For problem one part b. there are 4 ways the white ball
    Can be chosen at least 2 times.
    WWW
    WWR
    WRW
    RWW
    The combined probability is the sum of the probability of each of these.
    So P(www) = 1/4 * 1/4 * 1/4 = 1/64
    P(WWR) = P(WRW) = P(RWW) = 1/4 * 1/4 * 3/4 = 3/64
    So adding them up gives (3 * 3/64) + 1/64 = 10/64 = 5/32.
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  3. #3
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    Re: Probability Help

    Quote Originally Posted by linalg123 View Post
    For problem one part b. there are 4 ways the white ball
    Can be chosen at least 2 times.
    WWW
    WWR
    WRW
    RWW
    The combined probability is the sum of the probability of each of these.
    So P(www) = 1/4 * 1/4 * 1/4 = 1/64
    P(WWR) = P(WRW) = P(RWW) = 1/4 * 1/4 * 3/4 = 3/64
    So adding them up gives (3 * 3/64) + 1/64 = 10/64 = 5/32.
    Thank you so much mate.
    And thank you for the full proof explanation. I'm can't believe how I didn't think of that at the time. But it all seems logical now.

    Once again, thanks!
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