Probability question?

Question

King family agreed to go to five movies with one movie each day without any repetitions. Sadly, they have to walk to the movie theater. If it rains on the day of the movie, the movie trip is cancelled forever. There is a 20% probability of rain each movie day. How likely is it that the King's family will get to go to atleast 2 of the 5 movie trips.

Answer 1

Dear mathfreak,

You can answer this question by using the binomial distribution. Recall that with the binomial distribution, the probability of exactly k successes is given by

f(k) = C(n,k) p^k (1 - p)^(n - k),

for k in {0, 1, 2, . . . , n} (otherwise f(k) = 0),

where n is the number of trials, p is the probability of success on any particular trial, and C(n,k) = n! / [k! (n - k)!] (known as the binomial coefficient, which is the number of ways that exactly k items can be chosen from a set of n items).

For your question, n = 5 is the total number of movie days, k is the number of days the King family succeeds in going to a movie (i.e., the number of movie days on which it doesn't rain), and p = 4/5 is the probability that it does not rain on each movie day (with 1 - p = 1/5, or 20%, being the probability of rain on each of those same days). Thus, you can compute the probability of the King family successfully going to the movies on exactly k movie days by calculating f(k).

Since you want to know the probability that the King family goes to the movies at least two times, you can calculate and sum the values of f(k) for k >= 2, or calculate f(k) for k < 2 and then subtract from 1. Thus,

P(King family goes to at least 2 movies on 5 days)
= f(2) + f(3) + f(4) + f(5)
= 1 - [f(0) + f(1)].

Notice that there are fewer terms to evaluate in this problem if you subtract the probabilities of going to fewer than 2 movies from 1. Doing so may make your life easier, but both approaches are correct and give the same result. The calculations are as follows.

f(0) = C(5,0) (4/5)^0 (1/5)^5 = 1 / 3125 = 0.00032

and

f(1) = C(5,1) (4/5)^1 (1/5)^4 = 4 / 625 = 0.00640,

so P(King family goes to at least 2 movies on 5 days)
= 1 - [0.00032 + 0.00640]
= 1 - 0.00672
= 0.99328 .

Answer 2

Why is this difficult? You do not add the probabilities. Each day has a 20% chance of rain, so there is an 80% chance they will go to the movies - all of the movies.

Answer 3

64%

There is an 80% chance on the first day. The second day (having made the first day) there is an 80% chance. But you still need the first day, which was 80%. So the chances of making it the 2nd day is 80% of 80%, which is 64%.

The chances of the 3rd day would be 80% or 64%, and so on.