The probabilities of success in an examination for three students X, Y and Z are 1/5, 1/4 and 1/3 respectively. Assuming independence of their performances, what is the probability that at least two of them pass the examination?

Introduction / Context:
This question explores probability with independent events and combines them using the ideas of exactly k successes and at least k successes. We have three students, each with a different probability of passing an examination, and we want the probability that at least two pass.

Given Data / Assumptions:

• Probability that X passes, P(X) = 1/5.
• Probability that Y passes, P(Y) = 1/4.
• Probability that Z passes, P(Z) = 1/3.
• The events of X, Y and Z passing are independent.
• At least two pass means either exactly two pass or all three pass.

Concept / Approach:
We first compute the probabilities of exactly two students passing, considering each combination: X and Y only, X and Z only, Y and Z only. Then we compute the probability that all three pass. Adding these probabilities gives the probability that at least two pass. The complements that each student fails are needed as well and are 1 minus each success probability.

Step-by-Step Solution:
Let qX = 1 - P(X) = 4/5, qY = 1 - P(Y) = 3/4, qZ = 1 - P(Z) = 2/3.Probability that exactly X and Y pass, but Z fails: P(X) * P(Y) * qZ = (1/5) * (1/4) * (2/3) = 2 / 60 = 1 / 30.Probability that exactly X and Z pass, but Y fails: P(X) * P(Z) * qY = (1/5) * (1/3) * (3/4) = 3 / 60 = 1 / 20.Probability that exactly Y and Z pass, but X fails: P(Y) * P(Z) * qX = (1/4) * (1/3) * (4/5) = 4 / 60 = 1 / 15.Sum for exactly two successes = 1/30 + 1/20 + 1/15.Bring to common denominator 60: 1/30 = 2/60, 1/20 = 3/60, 1/15 = 4/60, sum = 9/60 = 3/20.Probability that all three pass = P(X) * P(Y) * P(Z) = (1/5) * (1/4) * (1/3) = 1 / 60.Therefore, probability that at least two pass = 3/20 + 1/60 = 9/60 + 1/60 = 10/60 = 1/6.

Verification / Alternative check:
We could also compute the complement probability that at most one passes. That is the sum of the probabilities that no one passes and that exactly one passes. Subtracting this from 1 would again give the probability that at least two pass. Performing that calculation produces 1/6 as well, which confirms our result from another viewpoint.

Why Other Options Are Wrong:
The fractions 1/4, 1/3 and 1/2 are larger than 1/6 and would correspond to overestimating the probability, usually because of including cases where only one person passes, or because of mistakenly adding probabilities that are not mutually exclusive. Only 1/6 correctly captures exactly the outcomes with two or three successes.

Common Pitfalls:
Some learners misinterpret "at least two" and include the case where exactly one student passes. Others forget to use the complements qX, qY and qZ when computing probabilities of exactly two successes, leading to incorrect products. Another problem is double counting the situation where all three pass by adding it multiple times. Being systematic with the cases and using independence carefully avoids these issues.

Final Answer:
The probability that at least two of X, Y and Z pass the examination is 1/6.