Let E be the event that a corn crop has an infestation of ear worms, and let B be the event that a corn crop has an infestation of corn borers. Suppose that P(E) = 0.24, P(B) = 0.16, and P(E and B) = 0.13. Find the probability that a corn crop has either an ear worm infestation, a corn borer infestation, or both.

The probability that a corn crop has either an ear worm infestation, a corn borer infestation

P(EUB) = 0.27

Step-by-step explanation:

Explanation:-

Addition theorem on probability:-

If S is a sample space, and E , F are any events in S then

P(EUF) = P(E) +P(F) -P(E n F)

Let 'E' be the event that a corn crop has an infestation of ear worms

let 'B' be the event that a corn crop has an infestation of corn bores

P(EUB) = P(E) +P(B) -P(E n B)

given P(E) = 0.24 and P(B) = 0.16 and P(E n B) =0.13

P(EUB) = P(E) +P(B) -P(E n B)

P(EUB) = 0.24 + 0.16 - 0.13

           = 0.27

The probability that a corn crop has either an ear worm infestation, a corn borer infestation

P(EUB)=0.27

hard

step-by-step explanation:

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.