suppose a certain study habit is known to be 80% effective in improving the performance of students,,if 5 students with poor performance are trained to acquire the habit,what is the probability that exactly 4 will have an improved performance? more than 3 improve?

To solve this problem, we can use the binomial probability formula. The binomial probability formula calculates the probability of getting a certain number of successes in a fixed number of independent Bernoulli trials (in this case, the improvement of students).

The binomial probability formula is as follows:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)

Where:
P(X = k) is the probability of getting exactly k successes,
n is the number of trials,
k is the number of successful trials,
p is the probability of success in a single trial, and
(1-p) is the probability of failure in a single trial.

In this problem,
n = 5 (since there are 5 students),
k = 4 (for the first question: exactly 4 improve),
k > 3 (for the second question: more than 3 improve), and
p = 0.8 (since the study habit is known to be 80% effective).

Let's calculate the probabilities:

1. Probability that exactly 4 will have improved performance:
P(X = 4) = (5C4) * 0.8^4 * (1-0.8)^(5-4)
= (5) * 0.8^4 * (0.2)^1
= 5 * 0.4096 * 0.2
= 0.4096

Therefore, the probability that exactly 4 students will have improved performance is 0.4096 or 40.96%.

2. Probability that more than 3 will have improved performance:
P(X > 3) = P(X = 4) + P(X = 5)
= 0.4096 + (5C5) * 0.8^5 * (1-0.8)^(5-5)
= 0.4096 + 1 * 0.8^5 * 0.2^0
= 0.4096 + 1 * 0.32768 * 1
= 0.4096 + 0.32768
= 0.73728

Therefore, the probability that more than 3 students will have improved performance is 0.73728 or 73.728%.