Question

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is:(a) $\frac{17}{3^{5}}$(b) $\frac{13}{3^{5}}$(c) $\frac{11}{3^{5}}$(d) $\frac{10}{3^{5}}$

Answer 1

Answer:

It should be (c) 11/3^5

Explanation:

We want at least 4 correct answers.

So, we can have two cases:

1) All 5 correct

Probability = (1/3)^5  [Only one choice is correct out of three for 5 questions]

2) 4 correct and 1 wrong.

For 4 correct, Probability = 1/3^4

For 1 wrong: 2/3

Probability = (1/3^4) * (2/3) = 2/3^5

If any one of the 5 questions might have gone wrong. So there are 5 possibilities for the single wrong answer:

So, Probability = (2/3^5) *5 = (10/3^5)

Final, both cases add:

Net Probability: 1/3^5 + 10/3^5 = 11/3^5

Step-by-step explanation: