Using a discrete probability distribution, it is found that:
a) There is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.
b) There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.
c) The expected value is of 1.3 lawns mowed on a randomly selected day.
<h3>What is the discrete probability distribution?</h3>
Researching the problem on the internet, it is found that the distribution for the number of lawns mowed on a randomly selected dayis given by:
Item a:
P(X = 2) = 0.3, hence, there is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.
Item b:

There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.
Item c:
The expected value of a discrete distribution is given by the <u>sum of each value multiplied by it's respective probability</u>, hence:
E(X) = 0(0.2) + 1(0.4) + 2(0.3) + 3(0.1) = 1.3.
The expected value is of 1.3 lawns mowed on a randomly selected day.
More can be learned about discrete probability distributions at brainly.com/question/24855677
 
        
             
        
        
        
When x is 100 y will be 122
        
             
        
        
        
Hey there! I'm happy to help!
First let's combine all of the terms with a.
2a+2a=4a
Now the ones with b.
3b-b=2b
And the ones with c.
-4c+2c=-2c
So, our simplified expression is 4a+2b-2c.
Have a wonderful day! :D