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

Answer:

1. The probability that the student will get exactly 6 correct answers is $\frac{1120}{19683}$.

2. The probability that the student will get more than 6 correct answers is $\frac{43}{2187}$.

Step-by-step explanation:

From the given information it is clear that

The total number of equations (n) = 10

The probability of selecting the correct answer (p)= $\frac{1}{3}$

The probability of selecting the incorrect answer (q)= $1-p=1-\frac{1}{3}=\frac{2}{3}$

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

$P(r=6)=^{10}C_6(\frac{1}{3})^6(\frac{2}{3})^{10-6}$

$P(r=6)=210(\frac{1}{3})^6(\frac{2}{3})^{4}=\frac{1120}{19683}$

Therefore the probability that the student will get exactly 6 correct answers is $\frac{1120}{19683}$.

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

$P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{10-7}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{10-8}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{10-9}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{10-10}$

$P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{3}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{2}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{1}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{0}$

$P(r>6)=120\times \frac{8}{59049}+45\times \frac{4}{59049}+10\times \frac{2}{59049}+1\times \frac{1}{59049}=\frac{43}{2187}$

Therefore the probability that the student will get more than 6 correct answers is $\frac{43}{2187}$.