Concept of Probability in math first level

Question

I am trying to teach myself the concepts of probability and I was wondering if this is correct. I am only 13 years old and did not learn this yet. I am just reading parts of a probability book to get ahead of something that I might learn later. Can someone help me to solve this. I know I got the concept, but I need help solving a problem like this. If I get this problem I know I can solve the other ones similar to it. Can someone help me with this.

Whenever Sara rents a movie, the probability that it is a horror movie is $0.57$. Of the next five movies she rents, determine the probability, to the nearest hundredth, that no more than two of these rentals are horror movies.

Work:

No more than two means $P(2)+P(1)+P(0)$

We know that the probability of it being a horror movie is $0.57$. Let us say that $P($Horror$)=$ $0.57$

$P(1)=5*0.57*0.43^4$

$P(2)=4*0.57*0.43^3$

$P(0)=3*0.57*0.43^2$

Adding these probabilities would give $0.59$

Answer 1

$P(H) = .57$ (the probability that it is horror)

$P(N) = 1-.57=.43$ (The probability that a movie is not horror)

She is going two rent 5 movies, and she is calculating the probability that no more than 2 of the movies are horror movies.

So, we have: $P(0)$ = the probability that exactly 0 of the movies are horror. $P(1)$ = the probability that exactly 1 of the movies are horror. $P(2)$ = the probability that exactly 2 of the movies are horror.

$P(0)=.43\times.43\times.43\times.43\times.43$ (the probability that it is not Horror 5 times in a row)

$P(1)=.57\times.43\times.43\times.43\times.43$ (the probability that the first movie is a horror movie and the other 4 are not.) But any of them could have been a horror, so we multiply this by 5. So, really

$P(1)=5\times(.57\times.43\times.43\times.43\times.43)$

$P(2)=10\times((.57\times.57\times.43\times.43\times.43)$ There are 10 ways to have two horror movies and 3 non-horror movies.

Now we add like you indicated $P(0)+P(1)+P(2)$ to find out the overall probability.

$P(0)=0.0147$

$P(1)=0.0974$

$P(2)=0.2583$

$Total=0.3704$

Rounding this off to the nearest hundredth, we get $0.37$

Answer 2

In short you can't add probabilities like that. So we are renting five movies, let's call them A,B,C,D,E. Each of these has a .57 chance of being horror and the picks are independant. Recall that the probability that all five are horror is $.57^5$ which is pretty small. So what you want is that 0 movies are horror OR 1 movie is horror OR 2 movies are horror. There are some tricks behind this, as a commentor mentioned above called binomial distribution. However, this might help with some of the logic. Since we are dealing with OR (above) this we can add because the events are mutually exclusive, that is if one happens the others can't. So first we want the chance that no movies are horror. Well the chance a movie is not horror is $1-.57 = .43$ So this would be $.43^5$. There are 5 different ways to have one movie be horror. it could be the first one, or the second one etc. And there is a $.43 * .57^4$ chance of that particular movie being horror. So we have $5*.43^4 * .57$. Finally, we have 10 different ways two movies can be horror so $10*.43^3*.57^2$. Now we sum up these probabilities. $.43^5 + 5 * .43^4*.57 + 10 * .43^3 * .57^2 = .37... $ The above is the logic behind a binomial distribution, but hopefully will give you the logic necessary to solve similar problems in the future. The faster way is to learn how to solve this problem in general.

Suppose you were renting 100 movie at wanted no more than 30 to be horror. That would be very difficult (annoying) to do with the above method, which is why there is a way to do it faster.