Mathematics, 10.03.2020 16:57 JOIful

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 20% of the incoming calls involve fax messages, and consider a sample of 20 incoming calls. (Round your answers to three decimal places.)(a) What is the probability that at most 7 of the calls involve a fax message?(b) What is the probability that exactly 7 of the calls involve a fax message?(c) What is the probability that at least 7 of the calls involve a fax message?(d) What is the probability that more than 7 of the calls involve a fax message?

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Answer from: amelvin41

a) 0.968 = 96.8% probability that at most 7 of the calls involve a fax message

b) 0.055 = 5.5% probability that exactly 7 of the calls involve a fax message

c) 0.913 = 91.3% probability that at least 7 of the calls involve a fax message

d) 0.032 = 3.2% probability that more than 7 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either there is a fax message, or there is not. The probability of a call being a fax is independent from other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

p = 0.2, n = 20

(a) What is the probability that at most 7 of the calls involve a fax message?

This is $P(X \leq 7)$ .

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.80)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.80)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.80)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.80)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.80)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.80)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.80)^{14} = 0.1091$

$P(X = 7) = C_{20,7}.(0.2)^{7}.(0.80)^{13} = 0.0545$

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 + 0.0545 = 0.968$

0.968 = 96.8% probability that at most 7 of the calls involve a fax message

(b) What is the probability that exactly 7 of the calls involve a fax message?

$P(X = 7) = C_{20,7}.(0.2)^{7}.(0.80)^{13} = 0.0545$

Rounded to three decimal places

0.055 = 5.5% probability that exactly 7 of the calls involve a fax message

Either less than 7 of the calls involve a fax message, or at least 7 does. The sum of the probabilities of these events is 1. So

$P(X < 7) + P(X \geq 7) = 1$

We want $P(X \geq 7)$ . So

$P(X \geq 7) = 1 - P(X < 7)$

In which

P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.913