Answer:
1/3
Explanation:
The simulation will say that Jeffrey collected three different toys if the numbers in a group of three are either 0 or 1, a number more than 1 and less than 6, and a number 6 or more.
The groups {3, 7, 1}, {5, 1, 7}, {9, 2, 0}, {3, 0, 6} and {4, 6, 0} all simulate collection of all three toys from three boxes of cereal. Of the 15 simulated sets of three boxes, that is a probability of 5/15, or 1/3.
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Additional comment
The simulated toy numbers can be grouped in different ways. For example, we could take a vertical column of numbers as representing the purchase of 3 boxes of cereal. Doing that, we find 3 different toys only for the simulated values {4, 0, 6}, the last digits in the 4th column. The probability from considering the simulation that way is 1/15.
From this group of 45 numbers, one could calculate P(car) = 10/45, P(bus) = 19/45, and P(airplane) = 16/45. These vary a little from the theoretical values. Overall, with these values, the probability of 3 different toys from 3 boxes is about 0.200, only slightly different from the theoretical probability of 0.192.