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What is the probability of getting at least one $3$ when rolling two standard ($6$-sided) dice?

I was thinking the answer would be $11$ out of $36$, but the textbook says it is $10$ out of $36$.

Please help.

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    $\begingroup$ The $\frac{10}{36}$ is the probability of exactly one $3$. As the problem is stated, the answer is $\frac{11}{36}$. The event "at least one $3$" includes two $3$'s. $\endgroup$ – André Nicolas Mar 24 '15 at 4:13
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Check the problem. The ways to throw at least 1 three are: $$\begin{align} &\big\{(3, x) \mid x\in \{1,2,4,5,6\}\big\}\cup\big\{(x, 3) \mid x\in \{1,2,4,5,6\}\big\}\;\cup\;\big\{(3, 3)\big\} \\[3ex] \mathsf P(\text{at least 1 three}) & = \mathsf P(\text{exactly 1 three})+\mathsf P(2\text{ threes}) \\[1ex] & = \dfrac{10}{36}+\frac{1}{36} \\[1ex] & = \frac {11}{36} \end{align}$$

So you seem to be correct, unless the problem actually asked for the probability of exactly 1 three.

Otherwise the textbooks given answer does not match the question asked.

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The correct answer is: $\frac {1}{6} \times \frac{5}{6} + \frac {5}{6} \times \frac{1}{6} + \frac {1}{6} \times \frac{1}{6} = \frac {11}{36}$

The test book is in error.

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