During a single day at radio station WMZH, the probability that a particular song is played is 23% . What is the probability that this song will be played on exactly 2 days out of 7 days ? Round your answer to the nearest thousandth.

To solve this problem, we can use the binomial probability formula. The binomial probability formula is given by:<br /><br />$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$<br /><br />where:<br />- $P(X = k)$ is the probability of getting exactly $k$ successes in $n$ trials<br />- $\binom{n}{k}$ is the number of ways to choose $k$ successes from $n$ trials<br />- $p$ is the probability of success in a single trial<br />- $(1-p)$ is the probability of failure in a single trial<br /><br />In this case, we want to find the probability that the song will be played on exactly 2 days out of 7 days. So, we have:<br />- $n = 7$ (the number of days)<br />- $k = 2$ (the number of days the song is played)<br />- $p = 0.23$ (the probability that the song is played on a single day)<br /><br />Plugging these values into the binomial probability formula, we get:<br /><br />$P(X = 2) = \binom{7}{2} \cdot 0.23^2 \cdot (1-0.23)^{7-2}$<br /><br />Calculating this expression, we get:<br /><br />$P(X = 2) = \frac{7!}{2!(7-2)!} \cdot 0.23^2 \cdot (1-0.23)^5$<br /><br />$P(X = 2) = \frac{7!}{2!5!} \cdot 0.23^2 \cdot (0.77)^5$<br /><br />$P(X = 2) = \frac{5040}{2 \cdot 120} \cdot 0.0529 \cdot 0.3087$<br /><br />$P(X = 2) = 21 \cdot 0.0529 \cdot 0.3087$<br /><br />$P(X = 2) = 0.341$<br /><br />Therefore, the probability that the song will be played on exactly 2 days out of 7 days is approximately 0.341, rounded to the nearest thousandth.