We can solve the problem by using the probability binomial distribution model. The formula is:
[tex]P(X=x)=(^n_x)p^xq^{n-x}[/tex]Recall that:
[tex](^n_x)=^nC_x[/tex]Given:
number of samples(n) = 8
x = 3
8% of all major bridges in that city will have ratings of 4 or below implying that the probability of a bridge having a rating of 4 or below is 0.08.
Hence,
[tex]\begin{gathered} p\text{ = 0.08} \\ q\text{ = 1-p } \\ q\text{ = 1-0.08} \\ q\text{ = 0.92} \end{gathered}[/tex]The probability that in a random sample of 8 major bridges in the city, at least 3 will have an inspection rating of 4 or below in 2020 would be:
[tex]\text{Probability of at least 3 = 1 - Probability of at most }2[/tex]Probability of at most 2 can be reduced to:
[tex]P(x\text{ }<\text{ 3) = P(x =2) + P(x = 1) + P(x = 0)}[/tex]Evaluating the expression, we have:
[tex]\begin{gathered} P(x\text{ }<3)=^8C_2(0.08)^2(0.92)^{8-2}+^8C_1(0.08)^1(0.92)^{8-1}+^8C_0(0.08)^0(0.92)^{8-0} \\ =\text{ 0.10866 + 0.35702 + }0.51322 \\ =\text{ 0.9789}0 \end{gathered}[/tex]The probability that at least 3 will have a rating of 4 and below:
[tex]\begin{gathered} P(x\text{ }\ge3)\text{ = 1 - P(x }<\text{ 3)} \\ =\text{ 1 - 0.97890} \\ =\text{ 0.02110} \end{gathered}[/tex]Answer:
0.02110