To solve these problems, we can model the situation using the binomial distribution, since we have a fixed number of independent trials (calls), each with two possible outcomes (lines occupied or not).
The parameters we have are:
- The probability of success (the lines are occupied) \( p = 0.4 \)
- The probability of failure (the lines are not occupied) \( q = 1 - p = 0.6 \)
- The number of trials (calls) \( n = 10 \)
(a) Probability that exactly three calls have the lines occupied
The probability mass function of a binomial distribution can be calculated using the formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where:
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)
- \( n = 10 \)
- \( k = 3 \)
- \( p = 0.4 \)
- \( q = 0.6 \)
Now, we calculate:
\[ P(X = 3) = \binom{10}{3} (0.4)^3 (0.6)^{10-3} \]
- Calculate \( \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120 \).
- Calculate \( (0.4)^3 = 0.064 \).
- Calculate \( (0.6)^{7} = 0.279936 \).
Now, substituting these values back into the formula:
\[ P(X = 3) = 120 \times 0.064 \times 0.279936 \]
Calculating this gives:
\[ P(X = 3) \approx 120 \times 0.064 \times 0.279936 \approx 2.016 \approx 0.2016 \]
So,
\[ \text{Probability that exactly three calls have the lines occupied} \approx 0.2016 \]
(b) Probability that at least one call has the lines not occupied
The probability that at least one call has the lines not occupied can be found by first calculating the probability that all calls have the lines occupied, and then subtracting that from 1.
The probability that all calls (10 calls) have the lines occupied is \( P(X = 10) \):
\[ P(X = 10) = \binom{10}{10} (0.4)^{10} (0.6)^{0} = 1 \times (0.4)^{10} \]
Calculating \( (0.4)^{10} \):
\[ (0.4)^{10} = 0.0001048576 \approx 0.0001 \]
Thus,
\[ P(\text{at least one not occupied}) = 1 - P(X = 10) = 1 - (0.4)^{10} \approx 1 - 0.0001 = 0.9999 \]
So,
\[ \text{Probability that at least one call has the lines not occupied} \approx 0.9999 \]
(c) Expected number of calls in which the lines are occupied
The expected value \( E(X) \) of a binomial distribution is given by:
\[ E(X) = n \cdot p \]
Substituting our values:
\[ E(X) = 10 \cdot 0.4 = 4 \]
So,
\[ \text{Expected number of calls in which the lines are all occupied} = 4 \]
Summary of Answers
(a) Probability that exactly three calls have the lines occupied: \( \approx 0.2016 \)
(b) Probability that at least one call has the lines not occupied: \( \approx 0.9999 \)
(c) Expected number of calls in which the lines are all occupied: \( 4 \)
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