Explain how this simple probability question has two answers

[pha0015]

Cambridge Senior Mathematical Methods Unit 3/4 Exercise 13B, question 5

The chance that a harvest is poorer than average is 0.5, but if it is known that a certain disease D is present, this probability increases to 0.8. The disease D is present in 30% of harvests. Find the probability that, when a harvest is observed to be poorer than average, the disease D is present.

Let H  be the event poor harvest.
Let  D  be the event disease.
Pr(D|H)=Pr(D∩H)/Pr(H)
=Pr(H|D)Pr(D)/(Pr(H|D′)Pr(D′)+Pr(H|D)Pr(D))
=0.8×0.3/(0.8×0.3+0.5×0.7)
=24/59

A second interpretation of the question which is entirely respectable.
Pr(D)=0.3
let  H  be a poor harvest
Pr(H)=0.5
Pr(H|D)=0.8
Pr(H∩D)=Pr(H|D)×Pr(D)
=0.8×0.3
=0.24
Pr(D|H)=Pr(H∩D)/Pr(H)
=0.24/0.5
=0.48

So the back of the book only showed the first answer, while I'd gotten the second one.

My question is how would it be possible for a probability question to have two different answers?
And why would there only be one answer in the back of the book, using the harder method (which I don't fully understand)?

[jazzycab]

Cambridge Senior Mathematical Methods Unit 3/4 Exercise 13B, question 5

The chance that a harvest is poorer than average is 0.5, but if it is known that a certain disease D is present, this probability increases to 0.8. The disease D is present in 30% of harvests. Find the probability that, when a harvest is observed to be poorer than average, the disease D is present.

Let H  be the event poor harvest.
Let  D  be the event disease.
Pr(D|H)=Pr(D∩H)/Pr(H)
=Pr(H|D)Pr(D)/(Pr(H|D′)Pr(D′)+Pr(H|D)Pr(D))
=0.8×0.3/(0.8×0.3+0.5×0.7)
=24/59

A second interpretation of the question which is entirely respectable.
Pr(D)=0.3
let  H  be a poor harvest
Pr(H)=0.5
Pr(H|D)=0.8
Pr(H∩D)=Pr(H|D)×Pr(D)
=0.8×0.3
=0.24
Pr(D|H)=Pr(H∩D)/Pr(H)
=0.24/0.5
=0.48

So the back of the book only showed the first answer, while I'd gotten the second one.

My question is how would it be possible for a probability question to have two different answers?
And why would there only be one answer in the back of the book, using the harder method (which I don't fully understand)?

From your calculation at the top, it seems the solutions have assumed that 0.5 is the probability that the harvest is poorer than average, given that the disease is not present. I, however, like you, did not read the question this way at all. The VCAA exam questions will certainly not be this ambiguous (that's not to say that you won't have to decipher them, but there will be something in the question that gives a clear indication as to whether a probability stated is conditional or not)