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Math Help - data management

  1. #1
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    data management

    It has been estimated that about 30% of beef contain bacteria to cause problem when partially cooked. Jane bought 15 beef. What is the probability that Jane will have less than 3 bacteria affected beef? Round to 2 decimal places
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  2. #2
    Senior Member
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    Hello

    Quote Originally Posted by tob2dam View Post
    It has been estimated that about 30% of beef contain bacteria to cause problem when partially cooked. Jane bought 15 beef. What is the probability that Jane will have less than 3 bacteria affected beef? Round to 2 decimal places
    p(beef contains bacteria) = 0.3 = 30%

    p(beef does not contain bacteria) = 1-0.3 = 0.7

    Now use the Binomial distribution

    {15 \choose 2}*0.3^2*0.7^{13}

    Jane bought 15 beefs, 2 are infected, therefore you need the propabilty: 1 beef is infected and 0 beef is infected

    {15 \choose 1}*0.3^1*0.7^{14}

    {15 \choose 0}*0.3^0*0.7^{15}

    p("less than 3 beefs are infected") = p("0 or 1 or 2 beefs are infected") = {15 \choose 0}*0.3^0*0.7^{15} + {15 \choose 1}*0.3^1*0.7^{14} + {15 \choose 2}*0.3^2*0.7^{13}
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