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In a multiple choice examination with three possible answers for each of the five questions out of which only one is correct, what is the probability that a candidate would get four or more correct answers just by guessing?
We have been given that, there are 3 possible answers of the total 5 questions out of which only 1 answer is correct.
Let p be the probability of getting a correct answer out of the 3 alternative answers.
Then, q is the probability of not getting a correct answer out of 3 alternatives.
And, p + q = 1
⇒ q = 1 – p
Let X be any random variable representing a number of correct answers just be guessing out of 5 questions.
Then, the probability that the candidate would get r answers correct by just guessing out of 5 questions is given by this Binomial distribution.
P (X = r) = nCrprqn-r
Here, n = 5.
Substitute the value of n, p, and q in the above formula.
…(i)
We need to find the probability that a candidate would get four or more correct answers just by guessing.
The probability can be expressed as,
Probability = P (X ≥ 4)
This is in turn can be written as,
P (X ≥ 4) = P (X = 4) + P (X = 5)
Put r = 4, 5 subsequently in equation (i) and then substitute in the above formula.
⇒ P (X ≥ 4) = 0.0453
Hence, the probability that a candidate would get four or more correct answers just by guessing is 0.0453.