It rains half the time in London and the weather forecast is correct in its prediction 2/3 of the time. You live in London and you always take an umbrella if the forecast predicts rain, and you never take an umbrella if the forecast does not predict rain. Calculate:
A) the probability that you are caught in the rain without an umbrella
B) the probability that you carried an umbrella on a day without rain
I used P(Rain) = P(Rain| Forecast Rain) * P(Forecast Rain) + P(Rain| Forecast no rain) * P(Forecast no rain).
Eventually I got P(Forecast Rain) = 1/2 and for A) I got 1/6 as an answer. Is this correct? I am pretty confused…
Best Answer
Here's one way:
A) Caught in the rain without an umbrella is the event (Rain and Forecast No Rain). Now,
$$P(\text{Rain and Forecast No Rain}) = P(\text{Forecast No Rain}|\text{Rain})\cdot P(\text{Rain}) = \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{6}$$
B) Carried an umbrella on a day without rain is the event (No Rain and Forecast Rain).
$$P(\text{No Rain and Forecast Rain}) = P(\text{Forecast Rain}|\text{No Rain})\cdot P(\text{No Rain}) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$
When in doubt with these kinds of problems, one thing you can always do is try to compute the conditional probabilities of the most specific events, ie, the 'and' events. Any more complicated event can be written as a disjoint union of the 'and' events, which is a sum for probabilities. So here, the specific events are (Rain and Forecast Rain), (Rain and Forecast No Rain), etc.