Answer:
The problem describes rolling an unfair number cube which lands with 6, and asks for the number of outcomes where X=0, X=1, X=2, and X=3.
X can be 0, 1, 2, or 3 if X equals the number of times she rolls a 6 over the course of three tries.
We must count the instances where none of the three trials yields a six in order to determine the number of occurrences where X = 0. Since there is a 0.4 percent chance that the cube will fall on 6, there is a 0.6 percent chance that it won't. Therefore, there are 0.6 * 3 = 0.216, or 21.6%, of outcomes where X = 0.
To find the number of outcomes where X=1, we need to count the number of outcomes where exactly one of the three trials results in a 6. There are three ways to choose which trial will result in a 6, and each of the other two trials must not result in a 6. Therefore, the number of outcomes where X=1 is 3 × 0.4 × 0.6^2 = 0.432 or 43.2%.
We must count the outcomes where precisely two out of the three trials yield a six in order to determine the number of events where X=2. There are three options for selecting the two trials that will end in a 6, and the third trial cannot also end in a 6. The proportion of outcomes where X=2 is therefore 3 0.4 2 0.6 = 0.288 or 28.8%.
Finally, to find the number of outcomes where X=3, we need to count the number of outcomes where all three trials result in a 6. This occurs with probability 0.4^3 = 0.064 or 6.4%.
Therefore, the number of outcomes where X = 0 is 21.6%, X=1 is 43.2%, X=2 is 28.8%, and X=3 is 6.4%.