# Pr radar detecting an aircraft

• September 17th 2009, 05:57 PM
Robb
Hi All,

I think I have the right answer for this, but I'm not that confident on these questions..

Three radar sets, operating independently, are set to tdetect any aircraft flying through a certain area. Each set has a probability of .02 of failing to detect a plane in its area. If an aircraft enters the area, what is the probability that it
a goes undeteected?
b is detected by all three radar sets?

So I defined the events;
$B_{1}$ - first fails
$B_{2}$ - second fails
$B_{3}$ - third fails.

Then $P(a)=P(B_{1} \cup B_{2} \cup B_{3})$ $=P(B_{1})+P(B_{2})+P(B_{3})$ $- \left[ P(B_{1} \cap B_{2} )+P(B_{1} \cap B_{3} )+P(B_{2} \cap B_{3} )\right] + \left[ P(B_{1} \cap B_{2} \cap B_{3})\right]$

With $P(B_{1} \cap B_{2})=P(B_{2} \cap B_{3} )=P(B_{2} \cap B_{3} )=0.02 \cdot 0.02 = 0.0004$
and $P(B_{1} \cap B_{2} \cap B_{3})=0.02 \cdot 0.02 \cdot 0.02 = 0.000008$
So $P(a) = 0.02+0.02+0.02-\left[ 0.0004+0.0004+0.0004\right] + 0.000008=0.058808$

And b detected by all three;
$P(b)=P(\bar{B_{1}} \cap \bar{B_{2}} \cap \bar{B_{3}}=0.98 \cdot 0.98 \cdot 0.98=0.941192$

Thanks for any help
• September 18th 2009, 08:59 AM
Pim
b is correct. However, a is not.

$P(B_{1} \cup B_{2} \cup B_{3})$
is the probability that any of the three radars does not detect it.
The probability that none detect it, would be:
$P(B_{1} \cap B_{2} \cap B_{3})$
As this means: 1 does not detect AND 2 does not detect AND 3 does not detect. And you worked that one out already.