Probability 1

• Feb 18th 2006, 04:46 AM
Natasha
Probability 1
A man is off work if either he or his wife is ill or both (he needs to stay at home to look after his children if his wife is ill). The probability that he is ill is 0.05, the probability that his wife is ill is 0.02, the probability that both are ill is 0.005.

1) What is the probability he is off work?

2) Find the conditional probability that he is ill given his wife is? Compare this with the corresponding unconditional probability: does it seem reasonable?

3) If you know he is off work, what is the probability he is ill?

• Feb 18th 2006, 08:13 AM
CaptainBlack
Quote:

Originally Posted by Natasha
A man is off work if either he or his wife is ill or both (he needs to stay at home to look after his children if his wife is ill). The probability that he is ill is 0.05, the probability that his wife is ill is 0.02, the probability that both are ill is 0.005.

1) What is the probability he is off work?

Let $MI$ denote the event that the man is ill, and $WI$ denote the event that his wife is ill.
Then the probability of the event that he is off work $OW$ is:

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

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)

Quote:

2) Find the conditional probability that he is ill given his wife is? Compare this with the corresponding unconditional probability: does it seem reasonable?
$
P(MI \wedge WI)= P(MI|WI)P(WI)
$

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

$
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)

Quote:

3) If you know he is off work, what is the probability he is ill?
(
In a thousand days he is off work for 65 days, of which he is ill on
50, so:

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

RonL
• Feb 18th 2006, 08:21 AM
Natasha
Quote:

Originally Posted by CaptainBlack
Let $MI$ denote the event that the man is ill, and $WI$ denote the event that his wife is ill.
Then the probability of the event that he is off work $OW$ is:

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

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)

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

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

$
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:

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

RonL

That's so clear thanks RonL :)