# Thread: data management

1. ## 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

2. Hello

Originally Posted by tob2dam
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

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

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

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

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

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