The question can be solved by using the Binomial Probability formula. This is given as
![P(X)=^nC_x.p^X.(1-p)^(n-X)](https://img.qammunity.org/2023/formulas/mathematics/college/gg0yrngst09r6l0uuzt97uxdq5f6c4blwq.png)
From the question, we have the following:
![\begin{gathered} n=3 \\ X=3 \\ p=0.65 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h6zmu3gn1jemvpsksqzov0tir08il7wyqa.png)
Inputting into the formula, we have
![\begin{gathered} P(X)=^3C_3*0.65^3*(1-0.65)^(3-3) \\ P(X)=(3!)/(3!(3-3)!)*0.65^3*0.35^0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ccvdwdgqxs99tptx5ikd8ulux18um8fv6k.png)
Solving further, we have
![\begin{gathered} P(X)=1*0.274625*1 \\ \therefore \\ P(X)=0.274625 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/a3pe2a57f451t1efon0bp84rvlx4j05d93.png)
Therefore, the probability, to three decimal places, is 0.275.