(a) The random variable X can be defined as the number of successful operations out of the five performed. (b) The random variable X follows a binomial distribution. This is because there are a fixed number of independent trials (5 operations), each trial has two possible outcomes (success or failure), and the probability of success is the same for each trial (80% or 0.8). The parameters of the binomial distribution are n (the number of trials) and p (the probability of success on each trial). In this case, n = 5 and p = 0.8. (c) The probability that exactly three operations are successful can be found using the formula for the binomial probability: P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k)) where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials. So, P(X = 3) = C(5, 3) * (0.8^3) * ((1-0.8)^(5-3)) = 10 * 0.512 * 0.04 = 0.2048 or 20.48%. (d) The probability that fewer than two operations are successful is the sum of the probabilities that exactly 0 and exactly 1 operation are successful. P(X 3) = P(X = 4) + P(X = 5) = [C(5, 4) * (0.8^4) * ((1-0.8)^(5-4))] + [C(5, 5) * (0.8^5) * ((1-0.8)^(5-5))] = [5 * 0.4096 * 0.2] + [1 * 0.32768 * 1] = 0.4096 + 0.32768 = 0.73728 or 73.728%.

Question

(a) The random variable X can be defined as the number of successful operations out of the five performed.

(b) The random variable X follows a binomial distribution. This is because there are a fixed number of independent trials (5 operations), each trial has two possible outcomes (success or failure), and the probability of success is the same for each trial (80% or 0.8). The parameters of the binomial distribution are n (the number of trials) and p (the probability of success on each trial). In this case, n = 5 and p = 0.8.

(c) The probability that exactly three operations are successful can be found using the formula for the binomial probability:

P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.

So, P(X = 3) = C(5, 3) * (0.8^3) * ((1-0.8)^(5-3)) = 10 * 0.512 * 0.04 = 0.2048 or 20.48%.

(d) The probability that fewer than two operations are successful is the sum of the probabilities that exactly 0 and exactly 1 operation are successful.

P(X < 2) = P(X = 0) + P(X = 1)

= [C(5, 0) * (0.8^0) * ((1-0.8)^(5-0))] + [C(5, 1) * (0.8^1) * ((1-0.8)^(5-1))]

= [1 * 1 * 0.00032] + [5 * 0.8 * 0.00064]

= 0.00032 + 0.00256 = 0.00288 or 0.288%.

(e) The probability that more than three operations are successful is the sum of the probabilities that exactly 4 and exactly 5 operations are successful.

P(X > 3) = P(X = 4) + P(X = 5)

= [C(5, 4) * (0.8^4) * ((1-0.8)^(5-4))] + [C(5, 5) * (0.8^5) * ((1-0.8)^(5-5))]

= [5 * 0.4096 * 0.2] + [1 * 0.32768 * 1]

= 0.4096 + 0.32768 = 0.73728 or 73.728%.

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