Question

suppose a multiple choice test has 20 questions with each question having 4 choices, only one of which is correct that is %25. suppose an unprepared student writes the answers, each time randomly picking on the 4 choices. what is the probability that the student will get
A) exactly 4 answers correct
B)All the answers incorrect
C) at least 8 correct
D) what is mean and the standard deviation 

Answer 1

There are 4 choices for each question.
Therefore the probability of guessing a correct answer is p = 1/4.
The probability of guessing an incorrect answer is q = 1 - p = 3/4.

The experiment is modeled by the binomial distribution.
There are 20 trials, so n=20
To obtain k successes in n trials, the probability is
$_{n}C_{k} \,p^{k}\,q^{n-k$
where
$_{n}C_{k} = \frac{n!}{k!\,(n-k)!}$

Part A.
To obtain 4 correct answers, the probability is
P(4 correct) = ₂₀C₄ (1/4)⁴ (3/4)¹⁶
                    = 0.19

Part B
Note that the probability of all answers incorrect = probability of getting all answers correct.
Therefore
P(all incorrect) = P(all correct)
= ₂₀C₂₀ (1/4)²⁰ (3/4)⁰
= (1/4)²⁰ = 9.095x10⁻¹³
= 0  (approximately)

Part C
To obtain 8 correct answers, the probability is
P(8 correct) = ₂₀C (1/4)⁸ (3/4)¹²
                    = 0.061

Part D
The mean is
m = np
    = 20*(1/4)
    = 5

The standard deviation is
s = √(npq)
  = √(20*0.25*0.75)
  = 1.937