Question

Exercise 2.64. On a test there are 20 true-or-false questions. For each problem the student (a) knows the answer with probability p,(b) thinks he knows the answer, but is wrong with probability q, (c) is aware of the fact that he does not know the answer with probability r. We assume that these happen independently for each question and that p+q+ r = 1. If the student doesn't know what to answer then he will guess by choosing true or false with probability 1/2-1/2. What is the probability that he will get the correct answer for at least 19 questions?

Answer 1

Answer:

$\left(p+\dfrac{p}{2}\right)^{19}\left(20-19p-\dfrac{19r}{2}\right)$

Step-by-step explanation:

Let the probability that he gets the correct answer be m.

If he gets the answer, then either he knows it (a probability of p) or he does not know it but guesses correctly (r × 1/2). Therefore,

$m = p + \dfrac{r}{2}$

Let the probability that he gets the incorrect answer be n. Then, by being complementary events

$n=1-m =1 - p - \dfrac{r}{2}$

This is a binomial or Bernoulli distribution since he gets the answer either correctly or incorrectly. The probability of at least 19 correct answers out of 20 is the sum of the probabilities of 19 or 20 correct answers.

$P(\gt19) = P(19) + P(20)$

$P(\gt19) = \binom{20}{19}m^{19}n^1 + \binom{20}{20}m^{20}$

$P(\gt19) = 20\left(p + \dfrac{r}{2}\right)^{19}\left(1 - p - \dfrac{r}{2}\right) + \left(p + \dfrac{r}{2}\right)^{20}$

$= \left( p + \dfrac{r}{2}\right)^{19}\left(20\left(1 - p - \dfrac{r}{2}\right)+p + \dfrac{r}{2}\right)$

$P(\gt19) = \left(p+\dfrac{p}{2}\right)^{19}\left(20-19p-\dfrac{19r}{2}\right)$