Question

assume that random guesses are made for eight multiple choice questions on a SAT test so that there are n=8 trials, each with probability of success (correct) given by p=0.20

a) prob of number x of correct answers is exactly 7

b) prob of numer x correct answers is at 4

c) prob of number x correct answers is fewer than 3

d) prob of number x corrwct answers is no more than 2

e) prob of no correct answers

Answer 1

Ans . This is binomial probability

n=8, p= 0..20

=> 1-p = 0.80

We find all probabilities using R:

# n=8,p=0.20
> # P(X=7)
> (choose(8,7)*(0.20)^7*(0.80)^(8-7))
[1] 0.00008192
> # P(X=4)
> (choose(8,4)*(0.20)^4*(0.80)^(8-4))
[1] 0.0458752
>
> # P(X<3)= P(X=0)+P(X=1)+P(X=2)
> (choose(8,0)*(0.20)^0*(0.80)^(8-0))+
+ (choose(8,1)*(0.20)^1*(0.80)^(8-1))+
+ (choose(8,2)*(0.20)^2*(0.80)^(8-2))
[1] 0.7969178
> # P(X>2)=P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)
> (choose(8,3)*(0.20)^3*(0.80)^(8-3))+
+ (choose(8,4)*(0.20)^4*(0.80)^(8-4))+
+ (choose(8,5)*(0.20)^5*(0.80)^(8-5))+
+ (choose(8,6)*(0.20)^6*(0.80)^(8-6))+
+ (choose(8,7)*(0.20)^7*(0.80)^(8-7))+
+ (choose(8,8)*(0.20)^8*(0.80)^(8-0))
[1] 0.2030801
> # P(X=0)
> (choose(12,0)*(0.20)^0*(0.80)^(8-0))
[1] 0.1677722

(a) prob of number x of correct answers is exactly 7 = 0.00008192

(b) prob of number x correct answers is at 4 = 0.0458752

(c) prob of number x correct answers is fewer than 3 = 0.7969178

(d) prob of number x correct answers is no more than 2 = 0.2030801

(e) prob of no correct answers = 0.1677722