To solve the problem, we need to use the concepts of probability and the binomial distribution.
1. Probability that only one of the two candidates passes the exam:
Let \( p = \frac{1}{5} \) be the probability that a candidate passes the exam and \( q = 1 - p = \frac{4}{5} \) be the probability that a candidate fails.
We want to find the probability that exactly one of the two candidates passes. This scenario can happen in two ways:
- The first candidate passes and the second fails.
- The first candidate fails and the second passes.
The formula for the probability of exactly one candidate passing (out of two candidates) is given by:
\[ P(\text{exactly one passes}) = P(\text{one passes, one fails}) + P(\text{one fails, one passes}) \]
Calculating this:
\[ P(\text{one passes, one fails}) = p \cdot q = \left( \frac{1}{5} \right) \cdot \left( \frac{4}{5} \right) = \frac{4}{25} \]
\[ P(\text{one fails, one passes}) = q \cdot p = \left( \frac{4}{5} \right) \cdot \left( \frac{1}{5} \right) = \frac{4}{25} \]
So, the total probability that exactly one of the two candidates passes is:
\[ P(\text{exactly one passes}) = \frac{4}{25} + \frac{4}{25} = \frac{8}{25} \]
2. Probability that all three candidates pass the examination:
To find the probability that all three candidates pass, we calculate:
\[ P(\text{all three pass}) = p^3 = \left( \frac{1}{5} \right)^3 = \frac{1}{125} \]
Summary of results:
- The probability that only one of the two candidates passes the examination is \( \frac{8}{25} \).
- The probability that all three candidates pass the examination is \( \frac{1}{125} \).
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