(a) between 10% and 48%

(b) between 0% and 38%

(c) anywhere between 0% and 100%.

We are not told that the events are independent. Arriving on time and arriving prepared seem like events that should have positive correlation - if he's too disorganized to get there on time, he's likely to be unprepared in other ways. This would give (a) and (b) higher probabilities than if the events were independent. But maybe they have negative correlation instead - he is late because he had to finish preparing.

How to get answer (a):

Consider 100 instances of Murray arriving for class. In 48 of them he is on time, and in 62 of them he is prepared.

It is possible that every class Murray arrives on time for, he is also prepared for - in which case he is both on time and prepared 48 out of 100 classes. (And prepared but late for 14 classes, and unprepared and late for 38 classes).

It is also possible that Murray is on time to all 38 classes for which he was unprepared. Which leaves 10 other classes for which he was on time, which must be ones he was prepared for.

The probability that Murray will get someone to do his homework for him is practically nil.

Technically we can't answer this because the two events (on time, prepared) may be dependent events. In other words, maybe on the days when he isn't prepared, he needs to take longer to get ready or is just more disorganized in general.

However, if we assume the two events are completely independent, then we can figure these out.

Event T = On Time

Event P = Prepared for Class

PART A:

P(T and P) = P(T) * P(P) = 0.48 * 0.62

Calculation left to the reader

PART B:

P(T' and P') = P(T') * P(P')

= (1 - P(T)) * (1 - P(P))

= 0.52 * 0.38

Calculation left to the reader

PART C:

If they are independent events, it shouldn't matter if he is prepared or not.

P(T|P') = P(T)

= 0.48