A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just guessing is

Options

A. `(17)/(3^5)`
B. `(13)/(3^5)`
C. `(11)/(3^5)`
D. `(1)/(3^5)`

Answer

C

Explanation

It is a problem on binomial distribution in which `n=5, p=` probability that the answer of a question is correct `=(1)/(3)and q=1-(1)/(3)=(2)/(3)`. Let X denote the number of correct answer. Then,
`P(X=r)=.^5C_(r)((1)/(3))^r((2)/(3))^(5-r), r=0,1,2,....,5`
Required Probability `=P(Xge 4)=P(X=4)+P(X=5)`
`=.^5C_(4)((1)/(3))^4((2)/(3))^1+.^5C_(5)((1)/(3))^5((2)/(3))^0=(10)/(3^5)+(1)/(3^5)=(11)/(3^5)`