Originally Posted by

**CaptainBlack** Let $\displaystyle MI$ denote the event that the man is ill, and $\displaystyle WI$ denote the event that his wife is ill.

Then the probability of the event that he is off work $\displaystyle OW$ is:

$\displaystyle P(OW)=P(MI)+P(WI)-P(MI \wedge WI)$$\displaystyle =0.05+0.02-0.005=0.065$,

this means that the probability is the sum of the probabilities of the

sub-events minus the probability that both sub-events occur (this is

because this is included within both sub-events will be double counted

if we do not subtract one of it).

Think of it like this: In 1000 days the man is ill on 50 days, his wife is ill

20 days, both are ill on 5 days, so he is off 50+20-5 days(this last to remove

double counted days)

$\displaystyle

P(MI \wedge WI)= P(MI|WI)P(WI)

$

where $\displaystyle P(A|B)$ denotes the probability that event $\displaystyle A$ occurs given that event $\displaystyle B$ occurs so:

$\displaystyle

P(MI|WI)=P(MI \wedge WI)/P(WI)=0.005/0.02=0.25

$

Which is reasonable. It tell us that if his wife is ill he is more likely

to be ill also.

(Of the 20 days in 1000 that his wife is ill he is also ill on 5 of the days)

In a thousand days he is off work for 65 days, of which he is ill on

50, so:

$\displaystyle

P(MI|OW)=50/65\approx 0.769

$

RonL