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A mail-order company business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. x 0 1 2 3 4 5 6 p(x) 0.12 0.15 0.20 0.25 0.18 0.07 0.03 Calculate the probability of each of the following events. (a) (at most three lines are in use} (b) [fewer than three lines are in use} (c) (at least three lines are in use}

A mail-order company business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. x 0 1 2 3 4 5 6 p(x) 0.12 0.15 0.20 0.25 0.18 0.07 0.03 Calculate the probability of each of the following events. (a) (at most three lines are in use} (b) [fewer than three lines are in use} (c) (at least three lines are in use}

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 32E
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### Probability Distribution of Telephone Lines in Use A mail-order company business has six telephone lines. Let \( X \) denote the number of lines in use at a specified time. The probability mass function (pmf) of \( X \) is provided in the accompanying table: | \( x \) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |:--------:|:---:|:---:|:---:|:---:|:---:|:---:|:---:| | \( P(X = x) \) | 0.12 | 0.15 | 0.20 | 0.25 | 0.18 | 0.07 | 0.03 | Calculate the probability of each of the following events: (a) **At most three lines are in use (≤ 3)**: \[ P(X \leq 3) \] (b) **Fewer than three lines are in use (< 3)**: \[ P(X < 3) \] (c) **At least three lines are in use (≥ 3)**: \[ P(X \geq 3) \] #### Solution: 1. **Probability of at most three lines in use:** \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \] \[ P(X \leq 3) = 0.12 + 0.15 + 0.20 + 0.25 = 0.72 \] 2. **Probability of fewer than three lines in use:** \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X < 3) = 0.12 + 0.15 + 0.20 = 0.47 \] 3. **Probability of at least three lines in use:** \[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \] \[ P(X \geq 3) = 0.25 + 0.18 + 0.07 + 0
Transcribed Image Text:### Probability Distribution of Telephone Lines in Use A mail-order company business has six telephone lines. Let \( X \) denote the number of lines in use at a specified time. The probability mass function (pmf) of \( X \) is provided in the accompanying table: | \( x \) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |:--------:|:---:|:---:|:---:|:---:|:---:|:---:|:---:| | \( P(X = x) \) | 0.12 | 0.15 | 0.20 | 0.25 | 0.18 | 0.07 | 0.03 | Calculate the probability of each of the following events: (a) **At most three lines are in use (≤ 3)**: \[ P(X \leq 3) \] (b) **Fewer than three lines are in use (< 3)**: \[ P(X < 3) \] (c) **At least three lines are in use (≥ 3)**: \[ P(X \geq 3) \] #### Solution: 1. **Probability of at most three lines in use:** \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \] \[ P(X \leq 3) = 0.12 + 0.15 + 0.20 + 0.25 = 0.72 \] 2. **Probability of fewer than three lines in use:** \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X < 3) = 0.12 + 0.15 + 0.20 = 0.47 \] 3. **Probability of at least three lines in use:** \[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \] \[ P(X \geq 3) = 0.25 + 0.18 + 0.07 + 0
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