Statistics Problems and Solutions Second Edition 2A.5 part (a)

Question

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So I was looking at Statistics Problems and Solutions (Second Edition) by Bassett et al., 1986, when I came across this question:

$2$A.$5\quad$Marking a multiple-choice test

(a) A multiple choice test paper contains$\,50\,$questions; for each question three answers are given, one of which is correct. The two incorrect answers to any one question are designed to be plausible, so that an ignorant candidate could be expected to pick an answer quite at random. If the examination is marked simply by giving one mark per correct answer, what should the pass mark be if the probability that a completely ignorant candidate passes is to be approximately $1$%?

My first attempt at solving (a) without looking at the answers:

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(a) We know that there is simply only $1$ mark given for getting a correct answer, and that there is a $1$ out of $3$ chance of getting any singular question correct. Therefore, we know the equation that we have to solve: $$50\left(\frac{1}{3}\right)^x\approx\frac{1}{100}$$$$\text{or}$$$$50\left(\frac{1}{3^x}\right)\approx\frac{1}{100}$$$$\implies\frac{\cancel{50}\left(\frac{1}{3^x}\right)}{\cancel{50}}\approx\frac{\left(\frac{1}{100}\right)}{50}$$$$\iff\frac{1}{3^x}\approx\frac{1}{5000}$$$$\implies\frac{1}{3^x}\approx0.002$$$$\text{which we can rewrite as}$$$$\log_35000\approx x$$$$\text{or}\,3^x\approx5000$$ Which we can solve like this: $$\log_{\cancel{3}}\cancel{3}^x\approx\log_35000$$ To get the approximate value of $x$, we need to find $\lceil3^{x}\rceil\approx5000$, which gets us $x=7.7526$.

Now here's the problem:

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That's obviously incorrect, since there's no way that there's a $1\%$ chance that $8$ out of $50$ are going to be correct.

Here's my $2$nd attempt at solve (a):

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$$\text{Since we know that of}\,\mathit{\text{any one answer that is chosen,}}$$$$\text{only one of them are going to be correct,}$$$$\text{so the question should be set up like this (using the normal distribution formula):}$$$$\mathit{X}\sim\operatorname{N}(x_0x_1,x_0x_1x_2)$$$$\to\mathit{X}\sim\operatorname{N}\left(50\cdot\frac{1}{3},50\cdot\frac{1}{3}\cdot\frac{2}{3}\right)$$$$\to\mathit{X}\sim\operatorname{N}\left(\frac{50}{3},\left(\frac{10}{3}\right)^2\right)$$ Where $$ \begin{align} \mathit{X}\, & \text{is defined as the number of questions needed to get right to pass} \\ \operatorname{N}(x_0x_1,x_0x_1x_2)\, & \text{is defined as the normal distribution formula} \\ x_0\, & \text{is defined as the number of questions} \\ x_1\, & \text{is defined as the chance of answering correctly on each question} \\ \text{and}\, x_2\, & \text{is defined as the chance of answering incorrectly on each question} \\ \end{align} $$ The problem

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I know we need to solve $x$ where $x$ is defined as being the pass mark that the number of correct answers $X$ is greater than or equal to the number of questions needed to get correct to pass the test. (i.e where $\operatorname{Pr}(\mathit{X}\geq x)\approx0.01$) My only problem being, I have no idea where to go from here.

To clarify

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1. While I am stuck on (a), I'm more than sure that likely that I would be able to solve (b) and (c) with the solution to part (a) (i.e. this isn't going to be a three-part question).
2. If you want to know what page this question is on (sorry it took me so long to get to that), it's on Page 67.
3. I probably shouldn't mention this, as it isn't necessarily needed, however, the reason I decided to omit parts (b) and (c) are because it isn't necessarily needed to understand part (a).

My question

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Would somebody be able to help me figure out where I went wrong/got stuck and how I could solve it more easily? I feel like I could easily solve this using my current knowledge, however I just don't know where to go from here.

Answer 1

The exact distribution of $X$ is given by a binomial distribution, with $X \sim B(100, \frac{1}{3})$. You're trying to find the threshold value $n$ for which $P(X \geq n) < 0.01$. This kind of calculation is fairly simple with modern computers, but for a question like this it's more likely that you're expected to use a normal approximation to the binomial. So in other words you want to take the value of $z$ for which $P(Z \geq z) \approx 0.01$ (Where $Z$ is a standard normal distribution), and a suitable transformation between $Z$ and $X$ so that $P(X \geq n) \approx P(Z \geq z)$.