Practice Final B6
tdoelker
I am having some trouble with the solution to this problem. I know it will involve complements and taking it to the 10th power and that p will = $\frac{1}{6}$ x $\frac{1}{10}$ I'm just confused as to how to piece that information together to get the right answer.
tdoelker: March 19, 2015, 4:35 p.m.
ACTUALLY THIS IS B7
Lilyhui: March 19, 2015, 5:15 p.m.
P= the probability of contracting the disease from one person on one exposure. P(contract) = 1-P(Dont contract) Here P= (1/6 x/10)= 1/60 You take the compliment to get the prob of not contracting the disease. 1-p= (59/60) Since you care coming into contact with 10 people, raise the prob to the power of 10 So then the final answer will be 1-(59/60)^10
weisbart: March 20, 2015, 1:23 a.m.
This is, of course, correct. The problem is a little harder if you are with the same person multiple times. Consider the following. The probability someone has a disease is $\frac{1}{4}$. The probability you get on a single exposure from someone who has it is $\frac{1}{10}$. The probability you get it from a random person on five exposures is \[P({\rm get\; disease}) = P({\rm get \; disease}|{\rm person\; has \;it})P({\rm person\; has\; it}) = \left(1 - \left(\frac{9}{10}\right)^5\right)\times \frac{1}{4}.\]