hii could i please have help on this question? im not sure what i did wrong

s JE AP Statistics Chapter 6 Take Home Test Name: It is estimated that 20 percent of all backseat passengers don't wear seat belts. In a random sample of - 7 5 backseat passengers, what is the probability that at least two don't wear seat belts? ST X (A) 1 - (0.2)5 n = 5 p = . 2 p ( x = 2 ) = 1 ( 3 ) ( 2)2 (.8) 3 (B) 1 - [(0.8) 5 + 5(0.2) (0.8) 4] (C))10(0.2)2(0.8)3 S ! rooms (D) 10(0.2)3(0.8)2 + 5(0.2)4(0.8) + (0.2)5 2 :3 ! ( . 2 ) 2 ( . 8 ) 3 (E) 10(0.2)2(0.8)3 + 5(0.2) (0.8)4 + (0.8) 5 10 ( .2 ) 2 ( . 8 ) 3 121. = , 205 A manufacturer knows that 15 percent of the widgets coming off the assembly lip defect. If an inspector keon