Question

The probability of a person over the age of 53 developing cancer is .06, and the probability that she/he develops heart problems is .08. If a patient over 53 is selected at random, what is the probability that the patient will develop:

A) None of these deseases?
B) At least one of the diseases?
C) Both of the diseases?

Answer 1

this is making the rather broad and probably incorrect assumption that these are independent yes?

Answer 2

For none 0.96*0.92
For At least 0.06+0.08
For both 0.06*0.08

Answer 3

not sure about this answer. think we need "both" first.

Answer 4

calculate both as roshan said.
\[.06\times .08=.0048\]

Answer 5

Yeap, even I am not sure about my answer

Answer 6

But I guess this is correct

Answer 7

not "at least one" means one or the other or both. so we add and then subtract intersection
\[.06+.08-.0048\]

Answer 8

if you just add \[.06+.08\] you are counting the intersection twice

Answer 9

trick to this question is you have to work from bottom up

Answer 10

first answer c then b then a

Answer 11

Yeap,It has been a while I have not done Stat

Answer 12

and now that we have the probability that you get at least one, we can compute the probability that you get none by subtracting from one

Answer 13

so whats the final answer? im confused

Answer 14

so answer to c is
\[.0048\] and answer to b is
\[.1352\] and answer to a is
\[.8648\]

Answer 15

@hscheide do you want to start again slow from the beginning?

Answer 16

no i think i got it though thank you :) i might have another problem for you :P

Answer 17

ok i think if you look above i wrote all the steps