Final answer:
To find the probability that in a selection of 5 jugs the first 2 are good and the last 3 are spoiled, we multiply the probabilities of each event happening in sequence.
Step-by-step explanation:
The probability question asks: If you randomly choose 5 jugs from 43 milk jugs with 11 being spoiled, what is the probability that the first 2 are good and the last 3 are spoiled? To solve this, we use the concept of dependent events and conditional probability.
There are 43 - 11 = 32 good milk jugs, and hence the probability that the first jug is good is 32/43. If the first jug chosen is good, there are now 31 good jugs left, and the probability that the second jug is also good is 31/42 (since there are now only 42 jugs left).
Similarly, after two good jugs are chosen, we have 11 spoiled jugs and 41 total jugs remaining. The probability that the next jug is spoiled is 11/41, then 10/40 for the fourth jug, and 9/39 for the fifth jug. Multiplying these probabilities gives the overall probability for this sequence of events.
Calculation: P(first 2 good, last 3 spoiled) = (32/43) × (31/42) × (11/41) × (10/40) × (9/39) = inclusive fraction after calculation.