monge1980 wrote:On a multiple choice test with 5 questions, and 3 answer choices per question, what is the probability of answering at least 4 question correctly if each answer is chosen randomly?
(A) 5/243
(B) 10/243
(C) 11/243
(D) 1/15
(E) 1/12

As per problem statement, to get at least 4 questions right we can have two scenarios ::

1. 4/5 questions are right
2. 5/5 questions are right

Getting 4/5 ==> [(1/3)^4 * (2/3)^1) 5C4 = 10/243
Getting 5/5 ==> (1/3)^5 =1/243

Adding the probabilities for both the scenarios = 10/243+1/243 = 11/243

Hence Answer is (C).

Getting 4/5 ==> [(1/3)^4 * (2/3)^1) 5C4 = 10/243

6983manish, could you explain the equation here?

bajwa2307 wrote:

Getting 4/5 ==> [(1/3)^4 * (2/3)^1) 5C4 = 10/243

6983manish, could you explain the equation here?

Here we are to find the probability of getting 4 correct answers out of 5 questions.
As each questions has 3 options and 1 correct answer out of these options.

Probability of correct answer while attempting one questions is 1/3.
For four correct answers its 1/3^4.

Similarly , probability of getting wrong answers is 2/3 , and for one question its 2/3^1.

As we are doing it for 5 questions , it will be [(1/3)^4 * (2/3)^1] = 2/243.

As we don't care which 4 we get right, there are 5C4 (5 choose 4) = 5 different ways we can get 4/5 correct. Therefore, the chance of getting 4/5 correct is 5 * 2/243 = 10/243.

Hope this helps.

Thanks, i was getting confused why you put 5C4 in the equation. My doubt is clear now. Thanks alot!

OA is (C) ... good explanation 6983manish