Likelihood of my friend being able to guess skittle taste I'm preparing for a data science interview, and here's a question I encountered during my preparation:

Your friend claims he can tell the five colors of skittles apart by
taste alone. The probability of a skittle being any particular color
is 1/5. You give your friend 3 skittles and he gets 2 correct. Should
you believe him? What if you give him 100 and he gets 40 correct?

I'm pretty sure the answer is yes, I should believe them in both instances. Here's my reasoning:
Let $X_i$ be an indicator random variable equalling $1$ if my friend is correct on the $i^{\text{th}}$ guess so that $E(X_i) = 1/5$ and $\text{Var}(X_i) = 4/25$.
The expected number of successful outcomes are 3/5 and 20, and the variance on three guesses is $12/25$, so guessing two correctly is more than two SD above the mean, and the variance on $100$ guesses is $16$, so they are just over one standard deviation above the mean.
I'm really not sure if my reasoning is correct, and I'd appreciate any sort of insight.
 A: Consider the first case with 2 out of 3 correct: Under the null
hypothesis that your friend is purely guessing, the number
correct is $X \sim \mathsf{Binom}(n=3, p=1/5).$ A test of the null hypothesis against the the alternative that $p > 1/5$ rejects for
large values of $X.$ So the P-value for outcome $X = 2$ is
$P(X \ge 2) = 0.104 > 0.05 = 5\%$ and you would not reject at the $5\%$ level. The evidence does not require you to believe
your friend can identify color by taste. [Computation below in R, but summing two terms using the binomial PDF is not difficult.
Note: If your friend got all three right, the probability of that just by guessing is $(1/5)^3 = 0.008$ and you should be convinced.]
sum(dbinom(2:3, 3, 1/5))
[1] 0.104

However, if your friend gets 40 out of 100 correct, then
the null distribution is $X \sim \mathsf{Binom}(n=100, p=1/5)$
and the P-value is $P(X \ge 40) \approx 0.$ So without
ability to judge color by taste, this outcome would be very rare.
You should believe your friend has some ability.
sum(dbinom(40:100, 40, 1/5))
[1] 1.099512e-28

By normal approximation to $\mathsf{Binom}(n=100, p=1/5),$
you have $\mu = E(X) = np = 20,\;$ $\sigma^2 =Var(X) = 16,\;$ $\sigma = SD(X) = 4.$ Then
$$P(X \ge 40) = P(X>39.5)\\ = P\left(\frac{X - \mu}{\sigma} > \frac{39.5-20}{4} = 4.875\right)\\ \approx P(Z > 4.875) \approx 0, $$
where $Z$ has a standard normal distribution.
1 - pnorm(4.875)
[1] 5.440423e-07

In the figure below, the P-value is the (very small) sum of heights of bars to the right of the vertical dotted line. The red curve shows the density function of the approximating normal distribution.

