Is it possible to pass a test all by chance?

Suppose you are taking a test of 100 questions, each of which has 4 options and values 1 point, and there is only 1 option is the correct answer. The pass line is set to 60 points. Now, you are holding a lottery of A, B, C, D to guess the correct answers all by chance.

Is it possible to pass the test all by chance?
Let’s think of it by serious mathematics.

First, because you try to choose a correct answer for each question, the possibility you succeed is \displaystyle p=\frac{1}{4}. And there is no doubt that the choosing action is a Bernoulli Experiment. So it is adapted to Binomial Distribution. Denote the random variable for a successful choice with \displaystyle X.

\displaystyle X \sim B(100,\frac{1}{4})

The probability we are chasing is:

\displaystyle P(X \geqslant 60)=?

In general, \displaystyle k successes in \displaystyle n choices,

\displaystyle P(X=k)\\\\=\binom{n}{k}p^k(1-p)^{n-k} \hspace{6mm} (k=0,1,2,..., n)

And, the cumulative distribution is:

\displaystyle \sum\limits_{k=0}^n \binom{n}{k}p^k(1-p)^{n-k}=1

So in our problem here,

\displaystyle P(X \geqslant 60)=\sum\limits_{k=60}^{100} \binom{100}{k}(\frac{1}{4})^k(\frac{3}{4})^{100-k}

Obviously, it is pretty hard to calculate this in that form.
But when \displaystyle n is big enough, we can transform the Binomial Distribution to Normal Distribution, by the famous “de Moivre–Laplace theorem“. So approximately,

\displaystyle X \sim N(np, np(1-p)).

By the way, a Normal Distribution density function is:

\displaystyle f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left({-\frac{(x-\mu)^2}{2\sigma^2}}\right).

\displaystyle F(X \geqslant x)=\int^{\infty}_x \frac{1}{\sqrt{2\pi}\sigma}\exp\left({-\frac{(x-\mu)^2}{2\sigma^2}}\right)dx.

And approximately, what we should do in our problem is, get the result of:

\displaystyle F(X \geqslant 60)\\\\=\int^{\infty}_{60} \frac{1}{\sqrt{2\pi}\sqrt{18.75}}\exp\left({-\frac{(x-25)^2}{2*18.75}}\right)dx

In order to look up in a Normal Distribution table, transform it to the Standard Normal Distribution format.

\displaystyle Z=\frac{X-\mu}{\sigma}=\frac{X-25}{\sqrt{18.75}}

Finally, use Standard Normal Distribution Table to find the result of this:

\displaystyle P(X \geqslant 60)\\\\=1-\Phi(Z \leqslant \frac{60-25}{\sqrt{18.75}})\\\\ \approx 1-\Phi(Z \leqslant 8.0829)\\\\ \approx 0

.
Ok, you see it, it is wise to study hard and do not play a lottery in your test. It does not help anyway.

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