Recall the binomial probability
[tex]\begin{gathered} P(x)=\binom{n}{x}\cdot p^x\cdot(1-p)^{n-x} \\ \text{where} \\ n\text{ is the sample size} \\ p\text{ is the probability of success} \end{gathered}[/tex]We have the following given
n = 7
p = 20% → 0.2
x = 6
Substitute these values and solve for the probability
[tex]\begin{gathered} P(x)=\binom{n}{x}\cdot p^x\cdot(1-p)^{n-x} \\ P(6)=\binom{7}{6}\cdot(0.2)^6\cdot(1-0.2)^{7-6} \\ P(6)=\binom{7}{6}\cdot(0.2)^6\cdot(0.8)^1 \\ P(6)=0.0003584 \end{gathered}[/tex]Therefore, the probability is 0.0003584.
