91Ó°ÊÓ

The binomial probability formula is given by: $$P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{(n-k)}$$ where \(P(X=k)\) is the probability of having \(k\) successes in \(n\) trials, \(\binom{n}{k}\) is the number of ways to choose \(k\) successes from \(n\) trials (calculated as \(n! / (k! \cdot (n-k)!)\)), \(p\) is the probability of success, and \((1-p)\) is the probability of failure.

All five operations are successful, so we need to find the probability of 5 successes out of 5 trials: $$P(X=5) = \binom{5}{5} \cdot 0.8^5 \cdot (1-0.8)^{(5-5)}$$ Simplify and calculate: $$P(X=5) = 1 \cdot 0.8^5 \cdot 1 = 0.8^5 \approx 0.32768$$ So, the probability that all five operations are successful is approximately \(32.768\%\).

Exactly four operations are successful, so we need to find the probability of 4 successes out of 5 trials: $$P(X=4) = \binom{5}{4} \cdot 0.8^4 \cdot (1-0.8)^{(5-4)}$$ Simplify and calculate: $$P(X=4) = 5 \cdot 0.8^4 \cdot 0.2^1 \approx 0.4096$$ So, the probability that exactly four operations are successful is approximately \(40.96\%\).

Less than two operations are successful, so we need to find the probability of 0 or 1 success out of 5 trials. Since these are independent events, we can calculate the probabilities separately and then add them together. Calculate the probability of 0 successes: $$P(X=0) = \binom{5}{0} \cdot 0.8^0 \cdot (1-0.8)^{(5-0)} = 1 \cdot 1 \cdot 0.2^5 \approx 0.00032$$ Calculate the probability of 1 success: $$P(X=1) = \binom{5}{1} \cdot 0.8^1 \cdot (1-0.8)^{(5-1)} = 5 \cdot 0.8^1 \cdot 0.2^4 \approx 0.0064$$ Now, add these probabilities together: $$P(X<2) = P(X=0) + P(X=1) \approx 0.00032 + 0.0064 \approx 0.00672$$ So, the probability that less than two operations are successful is approximately \(0.672\%\). In conclusion, the probabilities for each event are: a. All five operations are successful: \(32.768\%\) b. Exactly four operations are successful: \(40.96\%\) c. Less than two operations are successful: \(0.672\%\)