Probability of throwing two dice twice, and getting a $7$ on the first throw

Question

This one is pretty simple, but the textbook answer is way off from mine. Getting a $7$ on a $2$-dice throw is $\frac 6{36} =\frac16$, and since it doesn't care about the second throw it should be $\frac16$, right? The textbook says it's $\frac5{18}$, but I don't know from where.

Answer 1

Your answer is correct. There are $36$ different possible results from rolling two dice and six of which result in a sum of seven:

$$\{1,6\}, \{6,1\}, \{2,5\}, \{5,2\}, \{3,4\}, \{4,3\}$$

giving a probability of

$$p=\frac{6}{36}=\frac{1}{6}$$

As Reese pointed out, if the question asks for the probability of getting a sum of seven on the first or second toss, but not both, then we have a standard binomial:

$$\begin{align*} P(X=k)={n \choose k}p^k(1-p)^{n-k} &={2 \choose 1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\\\\ &=\frac{10}{36}\\\\ &=\frac{5}{18} \end{align*}$$