Pr(R|A)=0.99
Pr(R|A
c
)=0.1
The probability of false alarm (a false indication of aircraft presence)
=Pr(R∩A
c
)=Pr(R|
A
c
)Pr(A
c
)=0.1×0.95=0.095
.
The probability of missed detection (nothing registers, even though an aircraft is
present)
=Pr(R
c
∩A)=Pr(R
c
|A)Pr(A)=0.01×0.05=0.0005
.
The probability that aircraft is present given that radar registers its presence
=Pr(A|R)=Pr(R|
A)Pr(A)Pr(R|A)Pr(A)+Pr(R|A
c
)Pr(A
c
)=0.99×0.050.99×0.05+0.1×0.95≈0.3426
.
10.
On an average the number of hits on a certain web page are 20 hits per day. What is
the probability of observing more than 30 hits in a day?
Hint : Use Poisson Distribution
Ans: Let
?
denote the number of hits on a certain webpage in a day with average number of hits
equal to 20 i.e.
?(
?
)=20
. We can use Poisson distribution to model the uncertainty in such
cases i.e.
?
∼Pois(20)
. So the probability that
?
is more than
30
is
Pr(
?
>30)=
∑
i x
=
31
∞
20
x
e
−
20
x !
= 0.01347468
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- Summer '18
- Sagar Arora