Question

Take a guess: A student takes a multiple-choice test that has 9 questions. Each question has four choices. The student guesses randomly at each answer. Let X be the number of questions answered correctly. (a) Find P(4). (b) Find P (More than 2). Round the answers to at least four decimal places.

Answer 1

Answer:

a) P(4) = 0.1168

b) P(More than 2) = 0.3993

Step-by-step explanation:

For each question, there are only to possible outcomes. Either the students guesses the correct answer, or he guesses the wrong. Each question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

9 questions.

This means that $n = 9$

Each question has four choices.

This means that $p = \frac{1}{4} = 0.25$

(a) Find P(4).

This is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{9,4}.(0.25)^{4}.(0.75)^{5} = 0.1168$

So

P(4) = 0.1168

(b) Find P (More than 2).

Either the student answers 2 or less questions correctly, or the student answers more than 2 correctly. The sum of the probabilities of these events is 1. Then

$P(X \leq 2) + P(X > 2) = 1$

We want $P(X > 2)$. Then

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751$

$P(X = 1) = C_{9,1}.(0.25)^{1}.(0.75)^{8} = 0.2253$

$P(X = 2) = C_{9,2}.(0.25)^{2}.(0.75)^{7} = 0.3003$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0751 + 0.2253 + 0.3003 = 0.6007$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.6007 = 0.3993$

Then

P(More than 2) = 0.3993