- #1
mrmt
- 6
- 0
Hi Guys,
Long time reader first time poster...
This simple question has stumped me all day and I think I've finally cracked it! I'm hoping someone can confirm that, or tell me how wrong I am - either is fine :)
One in 1000 cows have a rare genetic disease. The disease is not contagious, therefore cases are independent.
Let X be the number of cows purchased by a farmer. How many cows are purchased by the farmer until the 1st cow with the disease, given:
P(X≤r)=0.05
P(X≤r)=0.90
This is what I've done:
p = 1/1000 = 0.001 (? was unsure if this was in fact my p value)
P(X>r)=(1-p)^r
P(X≤r)=0.05 (given)
P(X≤r) + P(X>r) = 1 for geometric distribution
Therefore:
0.05 + (1-p)^r=1
0.05 + (1-0.001)^r=1
0.999^r=0.95
ln(0.999^r)=ln(0.95)
r≈51
And same again for P(X≤r)=0.90
Can someone tell me if I'm heading in the right direction - or is there a better way?
Thanks
Long time reader first time poster...
This simple question has stumped me all day and I think I've finally cracked it! I'm hoping someone can confirm that, or tell me how wrong I am - either is fine :)
One in 1000 cows have a rare genetic disease. The disease is not contagious, therefore cases are independent.
Let X be the number of cows purchased by a farmer. How many cows are purchased by the farmer until the 1st cow with the disease, given:
P(X≤r)=0.05
P(X≤r)=0.90
This is what I've done:
p = 1/1000 = 0.001 (? was unsure if this was in fact my p value)
P(X>r)=(1-p)^r
P(X≤r)=0.05 (given)
P(X≤r) + P(X>r) = 1 for geometric distribution
Therefore:
0.05 + (1-p)^r=1
0.05 + (1-0.001)^r=1
0.999^r=0.95
ln(0.999^r)=ln(0.95)
r≈51
And same again for P(X≤r)=0.90
Can someone tell me if I'm heading in the right direction - or is there a better way?
Thanks