The probability a person in America will get Leukemia is 1/40,000. Consider a population of 200,000 people. Given one person has Leukemia, what is the probability that at least one other person has Leukemia?
I tried this using the law of total probability but kept getting answers greater than 1! Can someone please show me the right way to do it?!
2006-12-06 06:20:13 · 9 answers · asked by kate 1 in Science & Mathematics ➔ Mathematics
Carlsberg is probably the best lager in the world
2006-12-06 06:22:42 · answer #1 · answered by ROBSTER 4 · 0⤊ 3⤋
As far as I know (pretty sure), these answers are all wrong.
You could work out the probabilty of one other having it, two other having it etc.... and add them all together, but that would take quite a while :p
Really, you want to do:
1 - (probability nobody else has leukemia)
So
1 - (39,999/40,000)^199,999
Giving an answer of 0.99326 This makes sense - if only 1 in 200,000 had leukemia it would be quite unlikely.
This method is used because the question says "at least." If it was "just one other" it would be using normal methods.
2006-12-06 15:05:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
stick with simple arithmetic 1 in 40,000 2 in 80,000 3 in 120,000 just divide the total number by the probability ( 40,000 )
in your problem it is slightly harder ( very slightly ) you have 199,000 people with no Leukemia so divide this by 40,000 ( the fact that a person in your sample has Leukemia changes nothing - the SAMPLE that the odds are calculated from is not 40,000 but ALL THE PEOPLE IN AMERICA - 300,000,000 - SO you could choose a group of 10 and find that 3 had Leukemia)
ALSO note that the problem is not how many have Leukemia but how many OTHER than the one has Leukemia THAT is why you use 39,999 as the sample not 40,000
2006-12-06 14:24:12 · answer #3 · answered by Anonymous · 0⤊ 0⤋
well can't you just put it as a ratio?
1/40,000 * X/200,000 then just solve for X. X will then become the probability of the number of people to have leukemia
2006-12-06 14:28:49 · answer #4 · answered by gets flamed 5 · 0⤊ 0⤋
The probability of one person having the disease is 1/40,000
The probability for another person is also 1/40,000
The probability for two people selected at random, both having it is
1/40,000 x 1/40,000 = 1/1,600,000,000
= 1 in 1.6 billion
2006-12-06 19:29:06 · answer #5 · answered by Anonymous · 0⤊ 0⤋
em ye more than 1 can have it because maybe in the other ratio none of them would not have it.. im nly 14 so im nt tat helpful 4 dis sorta q bu i hope i helped u ou in someway xx
2006-12-06 14:22:59 · answer #6 · answered by xXxDolly92xXx 1 · 0⤊ 2⤋
i dont have a clue how to do it sorry i got 4 and 5
2006-12-06 14:23:30 · answer #7 · answered by conor210782 4 · 0⤊ 2⤋
I think there's a 50% chance you will get this question wrong
2006-12-06 14:23:17 · answer #8 · answered by Anonymous · 0⤊ 2⤋
4/200,0000
2006-12-06 14:27:36 · answer #9 · answered by Christopher McGregor 3 · 0⤊ 0⤋