[OZAPRS] HF Mobile beacon times

Chris Hill chris.hill at crhtelnet.com.au
Thu Jul 8 01:38:06 EST 2004

```Hi Richard,

>In the example above  VK3BYD-14 has a 33% chance of being heard by
VK4DMI-4
>and a 50% chance of being heard by VK4AH-4. So he has a 83% chance of
being
>heard. The WA station has a 50% chance and the NT station has a 33%
chance.

Just a fine point on probabilities...

If
the probability of being heard by station 1 = P1 = 50%
and
the probability of being heard by station 2 = P2 = 33%

Then
the probability of being heard by both stations = Pall = P1 * P2 = 0.5 *
0.33 = 0.165
and
the probability of not being heard at all = P! = P!1 * P!2 = 0.5 * 0.67 =
0.33
and
the probability of being heard by one or both stations = 1 - P! = 0.67

In other words, the station mentioned above has a 67% chance of being
heard,
not an 83% chance.

73,

Chris vk6kch

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