To solve this problem, we can use the binomial distribution, which is appropriate here since we have a fixed number of independent trials (calls), each with two possible outcomes (occupied or not occupied).
Let's define the parameters:
- Probability that the lines are occupied \( p = 0.4 \)
- Probability that the lines are not occupied \( q = 1 - p = 0.6 \)
- Number of calls \( n = 10 \)
We can use R programming to compute the required probabilities and the expected number of calls. Here are the R statements for each part:
(a) Probability that for exactly three calls, the lines are occupied
We will use the dbinom function, which calculates the probability of getting exactly \( k \) successes (occupied calls) in \( n \) trials.
# Parameters
n <- 10 # number of calls
p <- 0.4 # probability of occupied lines
k <- 3 # number of calls where the lines are occupied
# Probability of exactly k occupied calls
probability_exactly_3 <- dbinom(k, n, p)
probability_exactly_3
(b) Probability that for at least one call, the lines are not occupied
To find the probability that at least one call is not occupied, it is easier to calculate the complementary probability (that all lines are occupied) and subtract from 1.
# Probability that all lines are occupied (0 calls not occupied)
probability_all_occupied <- dbinom(n, n, p)
# Probability that at least one call is not occupied
probability_at_least_one_not_occupied <- 1 - probability_all_occupied
probability_at_least_one_not_occupied
(c) Expected number of calls in which the lines are occupied
The expected number of successes in a binomial distribution can be calculated using the formula \( E(X) = n \cdot p \).
# Expected number of occupied calls
expected_occupied <- n * p
expected_occupied
Summary
You would run the above R statements in an R environment to compute the probabilities and expected values for each part of the problem.
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