Probability to get 170 1's Suppose we throw a fair die 1000 times. What is the probability we get 170 1's?

The exact number is $P={1000\choose 170}\bigg(\frac{1}{6}\bigg)^{170}\bigg(\frac{5}{6}\bigg)^{830}$.

We can approximate with normal distribution because we have independent events, a large and constant number of trials and an a-priori known probability with only two outcomes, success (get an 1) or failure (get anything else).
We have $μ = np = 1000 \cdot\frac{1}{6} = 166.66$ and
$σ = \sqrt npq = 11.785$.
$z = \frac{170 - 166.66}{11.785} = 0.2828$
Hence the probability is 0.6103.
Is it correct? Thank you very much.
 A: No.
First, let's label each distribution so as to make clear which distribution I am referring to.
Let $X$ be the original, exact Binomial distribution: $X\sim B\left(1000,\frac{1}{6}\right).$ Then you want $P(X=170),$ and the calculation you suggest, $P={1000\choose 170}\bigg(\frac{1}{6}\bigg)^{170}\bigg(\frac{5}{6}\bigg)^{830}$, is correct, and gives you the exact probability of $P(X=170).$
Let's call the normal approximation to this distribution $Y,$ which is valid since $np$ and $np(1-p)$ are both $>5$. You should always check this fact before continuing with a normal approximation to a Binomial distribution. $Y\sim N\left(\frac{1000}{6}, \frac{5000}{36}\right),$ or roughly, $(167,12)$, although we use the exact values in calculations of course.
$Z$ is the standardised normal distribution: $Z\sim N(0,1)$ and note that when transferring between $Y$ and $Z$, all you are really doing is coding the mean and standard deviation of the distributions $Y$ and $Z$.
You have calculated $P(Z< 0.2828) = 0.6103,$ which is the same as $P(Y< 170),$ or equally, $=P(Y\leq 170)$. But this is the area to the left of $170$ on the $Y$ distribution (under the graph), so represents approximately (is equal to) $P(X\leq 170) = P(X=170) + P(X=169) + P(X=168)+\ldots.$ But the question asks for $P(X=170)$, and for this, you should calculate, $P(169.5<Y<170.5),$ and the "$0.5$ adjustment" is known as a "continuity correction". Although I'm not why this is considered "an adjustment".
A: No, the question is asking for P(X=170) and when you gave answer for normal approximation you found $P(X\leq 170)$. Use continuity correction, so when approximating to normal distribution calculate $P(170-0.5\leq X\leq 170+0.5)$.
