Respuesta :
The probability that the student gets between 1 and 6, exclusive, questions correct will be 0.7872...
Explanation
If the probability of success is [tex]p[/tex] , then the probability of [tex]r[/tex] successes in [tex]n[/tex] trials will be......
[tex]P = ^nC_{r}*(p)^r (1-p)^n^-^r[/tex]
Here, the quiz has total 12 questions and each question has 4 answer choices. So, [tex]n=12[/tex] and [tex]p=\frac{1}{4}= 0.25[/tex]
The student gets between 1 and 6, exclusive, questions correct. It means there can be 2 or 3 or 4 or 5 questions as correct. So....
The probability of 2 or 3 or 4 or 5 questions correct will be:
[tex]=^1^2C_{2}*(0.25)^2(1-0.25)^1^0 + ^1^2C_{3}*(0.25)^3(1-0.25)^9 +^1^2C_{4}*(0.25)^4(1-0.25)^8+^1^2C_{5}*(0.25)^5(1-0.25)^7\\ \\ =0.2322...+0.2581...+0.1935...+0.1032... \\ \\ =0.7872...[/tex]
