Question

4. Suppose that the next quiz contains 5 multiple choice questions with four choices for each questiorn (a, b, c, d). Unfortunately Esther forgot to study and chooses to randomly guess the answers What is the probability that a. The only question she gets right is the fifth question? b. She earns a satisfactory grade (considered, for this case, as an 80% or greater)? c. She gets no question right?

Answer 1

Question 4

Part a

Here, we have to find the probability that the only question she gets right is the fifth question.

There are total 5 questions.

Each question has 4 choices.

Out of these 4 choices, 1 choice is correct choice and 3 choices are wrong choices.

Probability of correct answer = P(C) = ¼ = 0.25

Probability of wrong answer = P(W) = ¾ = 0.75

We have to find the probability that fifth question is correct and others are incorrect or wrong.

That is,

Required probability = P(W)*P(W)*P(W)*P(W)*P(C)

Required probability = 0.75*0.75*0.75*0.75*0.25

Required probability = 0.079102

Part b

Here, we have to find the probability that she earns a satisfactory grade.

Total number of questions = n = 5

Probability for correct answer = p = 0.25

80% of 5 = 4

We have to find P(X≥4) given that n = 5, p = 0.25

We have to use binomial formula given as below:

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X≥4) = P(X=4) + P(X=5)

P(X=4) = 5C4*0.25^4*(1 – 0.25)^(5 – 4)

P(X=4) = 5*0.25^4*0.75^1

P(X=4) = 0.014648

P(X=5) = 5C5*0.25^5*0.75^0

P(X=5) = 1*0.25^5*1

P(X=5) = 0.000977

P(X≥4) = P(X=4) + P(X=5)

P(X≥4) = 0.014648 + 0.000977

P(X≥4) = 0.015625

Required probability = 0.015625

Part c

Here, we have to find P(X=0)

n = 5, p = 0.25

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.25^0*0.75^5

P(X=0) = 1*1* 0.237305

P(X=0) = 0.237305

Required probability = 0.237305