A problem on a multiple-choice quiz is answered correctly with probability 0.9 if a student is prepared. An unprepared student guesses between 4 possible answers, so the probability of choosing the right answer is 1/4. Seventy-five percent of students prepare for the quiz. If Mr. X gives a correct answer to this problem, what is the chance that he did not prepare for the quiz?

0.08475

Step-by-step explanation:

The question above is a application of conditional probability.

The formula to use is Baye's Theorem for conditional probability.

From the above question we have the following information:

Probability of answering correctly when prepared = 0.9

Probability of not answering correctly when prepared = 1 - 0.9 = 0.1

Probability of choosing the right answer = 1/4 = 0.25

Probability of choosing the wrong answer = 1 - 0.25 = 0.75

Number of students that prepare for the quiz = 75% = 0.75

Therefore number of students that did not prepare for the quiz = 1 - 0.75

= 0.25

Hence,

The probability of not preparing but choosing the correct answer =

P[ not prepared | correct answer ]

Is calculated as :

P[ not prepared | correct answer ] =

(0.25 × 0.25)/(0.25 × 0.25) + (0.25 × 0.9)

= 0.08475

Therefore, the chance that Mr X did not prepare for the quiz but he gives the right answer = 0.08475

answer:

step-by-step explanation:

the correct answer is x/3+10≥32

hope this !