Question

In a multiple choice quiz, there are 5 questions each with 4 answer choices (a, b, c, and d). Robin has not studied for the quiz at all and decides to randomly guess the answers.
A. What is the probability she gets the majority of the questions correct?
B. What is the probability the first question she gets right is the 3rd question?
C. What is the probability she gets exactly 3 or exactly 4 questions right?

Answer 1

Answer:

A.  0.1035

B.  0.1406

C.  0.1025

Step-by-step explanation:

Given that:

the number of sample questions (n) = 5

The probability of choosing the correct choice (p) = 1/4 = 0.25

Suppose X represents the number of question that are guessed correctly.

Then, the required probability that she gets the majority of her question correctly is:

P(X>2) = P(X=3) + P(X =4) + P(X = 5)

$P(X>2) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4} + (^{5}C_{5}) \times (0.25)^5 (1-0.25)^{5-5}$

$P(X>2) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} +\dfrac{5!}{5!(5-5)!} \times (0.25)^5 (1-0.25)^{0} \Bigg ]$

P(X>2) = [ 0.0879 + 0.0146 + 0.001 ]

P(X>2) = 0.1035

B.

Recall that

n = 5 and p = 0.25

The probability that the first Q. she gets right is the third question can be computed as:

$P(X=x) = 0.25 ( 1- 025) ^{x-1}$

Since, x = 3

$P(X = 3) = 0.25 ( 1- 0.25 ) ^{3-1}$

$P(X =3) = 0.25 (0.75)^{3-1}$

$P(X =3) = 0.25 (0.75)^{2}$

P(X=3) = 0.1406

C.

The probability she gets exactly 3 or exactly 4 questions right is as follows:

P(X. 3 or 4) = P(X =3) + P(X =4)

$P(X=3 \ or \ 4) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4}]$

$P(X=3 \ or \ 4) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} \Bigg ]$

P(X = 3 or 4) = [ 0.0879 + 0.0146 ]

P(X=3 or 4) = 0.1025