To find the probability that exactly 5 respondents out of 20 support the decision, we can use the binomial probability formula. The binomial probability formula is:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where

• $n$ is the total number of trials (in this case, 20),
• $k$ is the number of successful trials (in this case, 5),
• $p$ is the probability of success on a single trial (in this case, 0.6),
• $(1-p)$ is the probability of failure on a single trial,
• and $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

Substitute the given values into the formula:

$P(X = 5) = \binom{20}{5} (0.6)^5 (0.4)^{15}$

First, calculate the binomial coefficient:

$\binom{20}{5} = \frac{20!}{5!(20-5)!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15504$

Now calculate $(0.6)^5$ and $(0.4)^{15}$:

$(0.6)^5 = 0.6 \times 0.6 \times 0.6 \times 0.6 \times 0.6 = 0.07776$

$(0.4)^{15} = 0.4^{15} = 1.073741824 \times 10^{-6}$

Combine these calculations for the probability:

$P(X = 5) = 15504 \times 0.07776 \times 1.073741824 \times 10^{-6}$

$P(X = 5) = 15504 \times 0.07776 \times 0.000001073741824$

$P(X = 5) \approx 0.000428$

So, the probability that exactly 5 respondents support the decision is approximately 0.000428, or 0.0428%.