Question

A student takes a multiple-choice test that has 10 questions. Each question has four choices. The student guesses randomly at each answer. Let X be the number of questions the student gets correct. (a) Find P(X = 3). (b) Find P(X > 2). (c) To pass the test, the student must answer 7 or more questions correctly. Would it be unusual for the student to pass? Explain.

Answer 1

Answer:

1)0.2502

2)0.475

3)0.003505

Step-by-step explanation:

Total No. of question n= 10

There are four choices in each question

So, Probability of success $p = \frac{1}{4}$

Probability of failure q  = $1- \frac{1}{4}=\frac{3}{4}$

We will use binomial over here

$P(X=x)=^nC_r p^r q^{n-r}$

1)

$P(X = 3)=^{10}C_3 (\frac{1}{4})^3 (\frac{3}{4})^7\\P(X = 3)=\frac{10!}{3!7!} (\frac{1}{4})^3 (\frac{3}{4})^7\\P(X = 3)=0.2502$

2)  $P(X > 2)=1-P(X\leq 2)$

P(X>2)=1-(P(X=0)+P(X=1)+P(X=2))

$P(X>2)=1-(^{10}C_0 (\frac{1}{4})^0 (\frac{3}{4})^{10}+(^{10}C_1 (\frac{1}{4})^1 (\frac{3}{4})^9+^{10}C_2 (\frac{1}{4})^2 (\frac{3}{4})^8)$

$P(X>2)=1-((\frac{1}{4})^0 (\frac{3}{4})^{10}+(\frac{10!}{1!9!} (\frac{1}{4})^1 (\frac{3}{4})^9+\frac{10!}{2!8!} (\frac{1}{4})^2 (\frac{3}{4})^8)$

P(X>2)=0.475

3)

$P(X\geq 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\P(X\geq 7)=^{10}C_7 (\frac{1}{4})^7 (\frac{3}{4})^{3}+(^{10}C_8 (\frac{1}{4})^8 (\frac{3}{4})^2+^{10}C_9 (\frac{1}{4})^9 (\frac{3}{4})^1+^{10}C_{10} (\frac{1}{4})^{10} (\frac{3}{4})^0\\\\P(X\geq 7)=\frac{10!}{7!3!} (\frac{1}{4})^7 (\frac{3}{4})^{3}+\frac{10!}{8!2!} (\frac{1}{4})^8 (\frac{3}{4})^2+\frac{10!}{9!1!} (\frac{1}{4})^9 (\frac{3}{4})^1+\frac{10!}{10!0!}(\frac{1}{4})^{10} (\frac{3}{4})^0\\\\P(X\geq 7)=0.003505$