Find the probability of getting at least one 3 when 9 fair dice are thrown

"It is the case that at least one is a three" is equivalent to saying "It is not the case that none are a three"

Here, we use the axioms of probability that $Pr(A\cup B) = Pr(A)+Pr(B)$ when $A$ and $B$ are mutually exclusive events, that $A\cup A^c = \Omega$ the sample space, and that $Pr(\Omega)=1$ to conclude that $Pr(\text{At least 1 three}) = 1-Pr(\text{no threes})$

If you wanted to, you could calculate it directly as follows:

Break into cases based on the number of threes. "At least one three" is in this case equivalent to "Exactly one three or exactly two threes or exactly three threes or exactly four threes"

Each of these are disjoint events, so $Pr(\text{at least one three}) = Pr(\text{exactly one three})+Pr(\text{exactly two threes})+Pr(\text{exactly three threes})+Pr(\text{exactly four threes})$

We calculate each individually. In the case of exactly one three, pick which location the three occupies (4 choices) and pick the number for each other location (5 choices each). Divide by the sample space size. Alternatively, approach via binomial distribution. Either way, you will arrive at $Pr(\text{exactly one three})=4(\frac{1}{6})(\frac{5}{6})^3$

Similarly, we have the others as $\binom{4}{k}(\frac{1}{6})^k(\frac{5}{6})^{4-k}$ for the probability of getting exactly $k$ threes.

Adding, you have $Pr(\text{at least one three}) = \frac{4\cdot 5^3 + 6\cdot 5^2+4\cdot 5^1+4}{6^4}$, which after calculation you will find equals $1-(\frac{5}{6})^4$

The mistake of using $(\frac{1}{6})^4$ is that you calculated $Pr(\text{exactly four threes})$ which is only a small portion of the event at least one three.

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