You are planning to go on a picnic tomorrow at the state park... probability

You are planning to go on a picnic at the state park tomorrow. You will only go if it is not raining AND your significant other agrees to go with you AND the park's picnic pavilion is open. The weather forecaster is predicting a 40% chance of rain. Not only that but, your sweetheart is only 75% sure that he/she can go with you. The park ranger is 90% sure that the picnic pavilion will be open this time of year. Assuming independence of the events, what is the probability that you'll end up going on your picnic? Show your calculations.

Let N denote a random variable for the raining event and thus \bar{N} is the random variable of not raining event. Given that Pr(N) = 0.4 ---> Pr(\bar{N}) = 1- Pr(N) = 0.6.

Let A denote a random variable for the event of other agrees. Given that Pr(A) = 0.75.

Let O denote a random variable of park opening event, Given that Pr(O) = 0.9.

Finally, let G denote a random variable of going to the Park. Pr(G) is required.

Since the and can be interpreted as intersection (I will use \cup to denote intersection), thus,

Pr(G) = Pr(\bar{N} \cup A \cup O).

However, since the events are statistically independent, then,

Pr(G) = Pr(\bar{N} \cup A \cup O) = Pr(\bar{N}) * Pr(A) * Pr(O)
= 0.6 * 0.75 * 0.9 = 0.405.

Best Regards.

Use 'complement rule' and 'multiplication rule for independent events' . . .