If Tom throws the dart 30 times, the denominator of the fraction would be 30, as that is the total number, and if he actually hits the target 20 times, that would be the nominator of the fraction as that is the amount out of the total that he hits the target. So if the fraction is 20/30 for the times he hits the target, we can say that 10/30 is the fraction (or probability) that he will miss. We can simplify the fraction by dividing 10 on both sides, getting 1/3. On the bottom the question with the die has 3 scenarios, hit, miss, or neither. it misses when it is 5 or 6, it hits when the die is 1 or 4, and when the die is 2 or 3, nothing happens. as a die has 6 sides, the denominator is 6, and as there is 2 scenarios where you miss the numerator is 2. The fraction or probability for this one, would be 2/6 or 1/3. Using the same math for the following, you can get 1/3 for the die, 1/3 for the first spinner, 1/5 for the first number generator, 1/3 for the second number generator, and 1/3 for the second spinner. As the original part with Tom has a probability of 1/3, we are looking for equal probability, so the correct simulations would then be the die, the first spinner, the second number generator, and the second spinner, while the rest would be incorrect. Hope this helps!<span />
