A random number generator that draws digits at random with replacement from {0,1,2,3,4,5,6,7,8,9} is run three times. Find the chance that the digits drawn can form the number 105 by rearrangement if necessary. Start by filling in the blanks in the following sentence; the first two blanks should be filled with integers. "This experiment is like throwing ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ balls independently at random into ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ boxes labeled ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ." Then imagine where the balls must land to make the event occur.

This experiment is like throwing 3 balls independently at random into 3 boxes labeled 1, 2, 3.

The probability that the number 105 can be formed, with rearrengement if necessary, is 0.006

Step-by-step explanation:

The probability that a 105 (in that order) happens is equal to the probability of any other arrangement of 105 (in fact, every number is equally likely to be selected without rearrangement).

The total amount of rearrengements for 105 is 3! = 6 (the first digit can be anyone from 1, 0 or 5, for the second one we have only 2 possibilities and from the third one only one). The probability for a specific rearrengement to be picked is (1/10)³ = 1/1000, because we only have 1 favourable case over 10 for each of the 3 digits, and we need the 3 to be favourable, thus we need to power 1/10 by 3.

Therefore, the probability that a rearrengement of 105 is obtained is 6* 1/1000 = 3/500 = 0.006.

answer is : muchas/tipicas

have a nice day!

~recklesschicken