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?

Total number of questions n=5
Probability of correct answer p=1/3
Probability of incorrect answer q=2/3
Let X be the number of correct answers.
Then X follows a binomial distribution with parameters n=5 and p=1/3.
The probability mass function is given by:
P(X=k) = C(n,k) * p^k * q^(n-k)
where C(n,k) is the binomial coefficient, given by C(n,k) = n! / (k! * (n-k)!)
We want to find P(X≥4) = P(X=4) + P(X=5)
P(X=4) = C(5,4) * (1/3)^4 * (2/3)^(5-4) = 5 * (1/81) * (2/3) = 10/243
P(X=5) = C(5,5) * (1/3)^5 * (2/3)^(5-5) = 1 * (1/243) * 1 = 1/243
P(X≥4) = P(X=4) + P(X=5) = 10/243 + 1/243 = 11/243
Therefore, the probability that a student will get 4 or more correct answers just by guessing is 11/243.
This can be expressed as a fraction: 11/243
Converting to decimal: 11/243 ≈ 0.045267