1. A couple plans to have three children. The chance they have a girl is 0.8. The gender of one child is independent of the gender of another child. What is the probability of having one boy and two girls when having three children?

0.008
0.096

0.384

0.512

2. The table below shows the probability distribution for the number of videos rented. Notice the notation of x and P(x). What is the probability that a customer rents less than two videos?

X videos rented 0, 1, 2, 3, 4 or more
P(x) 0.1, blank, 0.3, 0.2, 0.1

3.

The table below shows the probability distribution for the number of videos rented. Notice the notation of x and P(x). What is the probability that a customer rents more than two videos?

X videos rented 0, 1, 2, 3, 4 or more

P(x) 0.1, blank, 0.3, 0.2, 0.1

4. The table below shows the probability distribution for the number of videos rented. Notice the notation of x and P(x). What is the probability that a customer rents less than three videos?

X videos rented 0, 1, 2, 3, 4 or more

P(x) 0.1, blank, 0.3, 0.2, 0.1

Number 1: To solve this one, is simple. All you have to do is know that, if there is a 0.8% possibility of having a girl, we will assume that it is also 0.8% chance to get a boy....

So... if we do 0.8^3 we would get 0.512

because you are having 3 children, and stating that you're predicting what the genders of all 3 children will be, you will cube the possibility to get the estimated chance of having 1 boy, and 2 girls...

Number 2: based on the table we know this, that there is a constant flow of numbers be added and subtracted...

from 0.3 to 0.2... its a 0.1 difference

from 0.2 to 0.2... its a 0.1 difference

by this information we can assume, that the table goes in order like this:

0.1, 0.2, 0.3, 0.2, 0.1

so it is a 0.2% possibility that a customer will rent less than 2 movies...

Number 3: The same answer that was given for Number 2, applies to this question....

Number 4: The same answer that was given for Number 2, applies to this question...