Question

A multiple choice test contains 10 questions, each with 3 possible answers (of which only one is correct). If a student answers each question by rolling a die and choosing the first answer if the die shows 1 or 2, the second answer if the die shows 3 or 4, or the third answer if the die shows 5 or 6, what is the probability that the student will get exactly 6 correct answers? more than 6 correct answers?

Answer 1

1. The probability that the student will get exactly 6 correct answers is
`(1120)/(19683)`.

2. The probability that the student will get more than 6 correct answers is
`(43)/(2187)`.

Explanation:

From the given information it is clear that

The total number of equations (n) = 10

The probability of selecting the correct answer (p)=
`(1)/(3)`

The probability of selecting the incorrect answer (q)=
`1-p=1-(1)/(3)=(2)/(3)`

According to the binomial distribution, the probability of selecting r items from n items is

`P=^nC_rp^rq^(n-r)`

where, p is probability of success and q is the probability of failure.

The probability that the student will get exactly 6 correct answers is

`P(r=6)=^(10)C_6((1)/(3))^6((2)/(3))^(10-6)`

`P(r=6)=210((1)/(3))^6((2)/(3))^(4)=(1120)/(19683)`

Therefore the probability that the student will get exactly 6 correct answers is
`(1120)/(19683)`.

The probability that the student will get more than 6 correct answers is

`P(r>6)=^(10)C_7((1)/(3))^7((2)/(3))^(10-7)+^(10)C_8((1)/(3))^8((2)/(3))^(10-8)+^(10)C_9((1)/(3))^9((2)/(3))^(10-9)+^(10)C_(10)((1)/(3))^(10)((2)/(3))^(10-10)`

`P(r>6)=^(10)C_7((1)/(3))^7((2)/(3))^(3)+^(10)C_8((1)/(3))^8((2)/(3))^(2)+^(10)C_9((1)/(3))^9((2)/(3))^(1)+^(10)C_(10)((1)/(3))^(10)((2)/(3))^(0)`

`P(r>6)=120* (8)/(59049)+45* (4)/(59049)+10* (2)/(59049)+1* (1)/(59049)=(43)/(2187)`

Therefore the probability that the student will get more than 6 correct answers is
`(43)/(2187)`.