$E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P\{X=3\}$ can be?

Question

Since each trial has the same probability of success, $p$, can you not uniquely solve for $p$? I.e:

Let, $X_{i} = 1$ if the $i^{th}$ trial is a success ($0$ otherwise). Then, $X=\sum_{i=1}^{3}X_{i}$, and $E[X] = E[\sum_{i=1}^{3}X_{i}] = \sum_{i=1}^{3}E[X_{i}] = \sum_{i=1}^{3}p =3p =1.8$

So, $p=0.6$, and $P\{X=3\}=0.6^{3}$

I thought what I did was sound, but the textbook says the answer to (a) is $0.6$ and (b) is $0$.

Their reasoning (for (a)) is as follows:

However, how can the above be true if all three trials have the same probability of success? That is, how can $P\{X=1\}$ and $P\{X=2\}$ be zero, when $P\{X=3\}$ is nonzero?

Answer 1

Hint: Nothing in the problem said that the trials are independent.