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Calculation of conditional probability (hail)
- Thread starter jones123
- Start date
\(\displaystyle \text{one has to assume some probability for hail if there is no thunderstorm. Once done you can apply}\)
\(\displaystyle P(H) = P(H|T) P(T) + P(H|!T)P(!T) \\ \text{where $T$ is the event a thunderstorm occurs and $H$ is the event hail occurs}\)
\(\displaystyle P(H) = P(H|T) P(T) + P(H|!T)P(!T) \\ \text{where $T$ is the event a thunderstorm occurs and $H$ is the event hail occurs}\)
Obiously, the chance of hail is zero if no thunderstorm occurs. This reduces the equation to:\(\displaystyle \text{one has to assume some probability for hail if there is no thunderstorm. Once done you can apply}\)
\(\displaystyle P(H) = P(H|T) P(T) + P(H|!T)P(!T) \\ \text{where $T$ is the event a thunderstorm occurs and $H$ is the event hail occurs}\)
P(H) = P(H|T) P(T) + P(H|!T)P(!T) = 0.25*0.5 = 0.125
Is that correct?
Thanks!
Well, it is not obvious that hail is caused exclusively by thunderstorms. I had to look up "hail" at wikipedia to confirm that particular meteorological fact. Perhaps you should have posed your question at a help site for meteorologists.
However, given that fact, the second summand in romsek's formula reduces to zero because
[MATH]\text {P(H given not T) * P(not T)} = 0 \text { * P(not T)} = 0.[/MATH]
So you are correct.
However, given that fact, the second summand in romsek's formula reduces to zero because
[MATH]\text {P(H given not T) * P(not T)} = 0 \text { * P(not T)} = 0.[/MATH]
So you are correct.