Solution
Part a
For this case we can do the following:
x 1 2 4 5
P(x) 1/10 5/10 2/10 2/10
P(x) 0.1 0.5 0.2 0.2
Part b
For this case the answer would be 0.2
Part c
For this case the answer would be.
P(X>1)= 0.5+ 0.2+0.2= 0.9
Part d
For this case we can find the expected value with this formula:
E(x) = 1*0.1 +2*0.5 +4*0.2+ 5*0.2= 2.9
Then we can conclude that the expected number of touchdowns for this case are 2.9