It is given that,
[tex]\begin{gathered} p\text{ = 8\% = 0.08} \\ n\text{ = 3} \end{gathered}[/tex]By applying Binomial distribution,
[tex]\begin{gathered} P(x\ge1)\text{ = 1 - P\lparen x}<1) \\ P(x\ge1)\text{ = 1- P \lparen x = 0 \rparen} \end{gathered}[/tex]Substituting the value in the formula,
[tex]\begin{gathered} P(x\ge1)\text{ = 1 - }^3C_x\left(0.08\right)^x\left(1-0.08\right)^{3-x} \\ P(x\ge1)=\text{ 1 - }^3\text{C}_x\left(0.08\right)^x\left(0.92\right)^{3-x}\text{ \_\_\_\_\_\_\_\_ x = 0, 1, 2, 3} \end{gathered}[/tex][tex]\begin{gathered} P(x\ge1)\text{ = 1 - 1}\times\text{ \lparen0.08\rparen}^0\times\left(0.92\right)^3 \\ P(x\ge1)\text{ = 1 - 1}\times\text{ 1 }\times0.7787 \\ P(x\ge1)\text{ = 1 - 0.7787} \\ P(x\ge1)\text{ = 0.221} \end{gathered}[/tex]Thus the required probability is 0.221.
