Amanda hits the target twice as often as she misses. What is

The first sentence of the question gives us the respective probabilities of Amanda hitting and missing the target.
Let the probability that Amanda hits the target be H and the probability that she misses the target be M. Then, as per the question, H = 2M.

But, Probability of Hitting + Probability of Missing = Total Probability. Therefore,

H + M = 1. Substituting the value of H in this equation, we have 3M = 1 or M = 1/3 and hence H = 2/3.

Of 5 trials, we want her to hit the target at least 4 times. This means we need to consider the following cases:

Case 1 – Amanda hits 4 times and misses once.
Probability = \((\frac{2}{3})^4\) * (\(\frac{1}{3}\)) * \(\frac{5! }{ 4!}\) = (\(\frac{16}{243}\)) * 5 = \(\frac{80}{243}\).

At this stage, since we know that the denominator is 243, we can eliminate options A, B and E since their denominator can never be 243 or a factor of 243. Knowing that there is one more case to be considered, we can also say that option C will not be the answer since we still have to add some value to 80/243.

Case 2 – Amanda hits 5 times and misses none.
Probability = \((\frac{2}{3})^5\) *\((\frac{1}{3})^0\) * \(\frac{5!}{5!}\) = \(\frac{32 }{ 243}\).

Therefore, total probability = \(\frac{80}{243}\) + \(\frac{32}{243}\) = \(\frac{112}{243}\).

The correct answer option is D. Hope that helps!