The machines in a warehouse can have zero, exactly one or 2+ defects with probablities 0.8, 0.15, and 0.05 respectively. Let D be the number of days till the machine fails. If the machine has zero defects then D~Geometric(p=0.001), if the machine has one defect then D~Geometric(p=0.005) and finally if the machine has 2+ defects, then D~Geometric(p=0.01) (a) What is the probability that a randomly chosen machine will fail after 79 days? (b) Suppose I give you a machine with one defect and let E be the event that the machine has not failed for 79 days. Given this information, what is the probability that D %26gt; 140? (c) Suppose I choose a random machine and let E be the event that the machine has not failed for 79 days. Given this information, what is the probability that D%26gt;140?
Stats and probability....please help!!?
Let L be life
we have several conditional probabilities
P(L%26lt;d|0 defects)=0.001*(1-.001)^(d-1)=0.001*(0.9...
P(L%26lt;d|1 defect)=0.005*(0.995)^(d-1)
P(L%26lt;d|2+ defects)=0.01*(0.99)^(d-1)
and
P(0 defects)=0.8
P(1 defect)=0.15
P(0 defects)=0.05
fail after 79 days = 1- Prob(failing 78 days and earlier)
P(L%26lt;d)=P(L%26lt;d|0 defects)P(0 defects)+P(L%26lt;d|1 defect)P(1 defect)+P(L%26lt;d|2+ defects)P(0 defects)
determine for d=78 and the subtract from 1
P(L%26gt;79)=1-P(L%26lt;78)=0.8*0.001*(0.999)^(7...
0.15*0.005*(0.995)^(78-1)+
0.05*0.01*(0.99)^(78-1)=
1-1.4811E-3=0.9985
I think this is a memoryless process. If I'm correct, it doesn't care that it's lasted 79 days. You need to find the probability it lasts another 69 days. Same with the third part -- you can use the function above.
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment