The probability a person selected at random has had exactly one visit to a doctor in the previous year is .45. Two patients are selected at random. Find the probability that both patients have had exactly one visit to a doctor.
2007-09-22 16:10:36 · 5 answers · asked by gibsonb2005 1 in Science & Mathematics ➔ Mathematics
I am not sure about the random part. If it is a set pool of people...then you are saying that 45 out of 100 have been to a doctor.
So the first person would be:
45/100=0.45
If there were only a 100 people in the random pool then your next person would have a 44/99 chance of having been to the doctor. So the chances would be:
45/100*44/99=0.20 probability.
However if it was a endless pool of people it would get closer to being 45/100*45/100=0.2025 probability. If you didn't know how many people were in the pool you'd go with this second one, but it would round to 0.20 anyways.
2007-09-22 16:18:07 · answer #1 · answered by Patty C 3 · 0⤊ 0⤋
Hello,
.45*.45 =.2025
Hope This Helps!!
2007-09-22 23:17:27 · answer #2 · answered by CipherMan 5 · 0⤊ 0⤋
probability of the first person visited the doctor is .45
probability of the second person visitted a doctor is also .45
P = (.45)^2
P = .2025 or 20.25%
2007-09-22 23:17:13 · answer #3 · answered by 7 · 0⤊ 0⤋
P = .45
both patients have had exactly one visit
P^2 = (.45)*(.45) = .2025 = 20.25%
2007-09-22 23:14:19 · answer #4 · answered by Anonymous · 1⤊ 0⤋
For any two independent events A and B:
P(A and B) = P(A) * P(B)
P(both have had exactly one visit to a doctor)
= 0.45 * 0.45 = 0.2025
2007-09-24 13:38:55 · answer #5 · answered by Merlyn 7 · 0⤊ 0⤋