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A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. What is the probability that a. At most 6 of the calls involve a fax message? b. Exactly 6 of the calls involve a fax message? c. At least 6 of the calls involve a fax message? d. More than 6 of the calls involve a fax message?

Sagot :

Step 1

The distribution given is a binomial distribution with following data:

[tex]n=25, \quad p=0.25[/tex]

Step 2:

(a) Probability of at most 6 calls involving a fax message:

We have to add probabilities for x=0 to x=6

[tex]\begin{aligned}& P(X \leq 6)=\sum_{x=0}^6{ }^{25} C_x(0.25)^x(1-0.25)^{25-x} \\& =\left(\begin{array}{l}{ }^{25} C_0(0.25)^0(1-0.25)^{25-0}+{ }^{25} C_1(0.25)^1(1-0.25)^{25-1} \\+{ }^{25} C_2(0.25)^2(1-0.25)^{25-2}+{ }^{25} C_3(0.25)^3(1-0.25)^{25-3} \\+{ }^{25} C_4(0.25)^4(1-0.25)^{25-4}+{ }^{25} C_{5 x}(0.25)^5(1-0.25)^{25-5} \\+{ }^{25} C_6(0.25)^6(1-0.25)^{25-6}\end{array}\right)\end{aligned}[/tex]

[tex]\begin{aligned}& =\left(\begin{array}{l}{ }^{25} C_0(0.25)^0(0.75)^{25-0}+{ }^{25} C_1(0.25)^1(0.75)^{24} \\+{ }^{25} C_2(0.25)^2(0.75)^{23}+{ }^{25} C_3(0.25)^3(0.75)^{22} \\+{ }^{25} C_4(0.25)^4(0.75)^{21}+{ }^{25} C_{5 x}(0.25)^5(0.75)^{20} \\+{ }^{25} C_6(0.25)^6(0.75)^{19}\end{array}\right) \\& =\left(\begin{array}{l}0.0007525+0.006271+0.02508+0.06410 \\+0.1175+0.003171+0.182819\end{array}\right) \\& =0.5611\end{aligned}[/tex]

Therefore,

[tex]P(X \leq 6)=0.5611[/tex]

Step 3:

(b) Probability of exactly 6 calls involving a fax message:

We have to find probabilities for x=6[tex]\begin{aligned}& P(X=6)={ }^{25} C_6(0.25)^6(1-0.25)^{25-6} \\& ={ }^{25} C_6(0.25)^6(0.75)^{19} \\& =0.182819\end{aligned}[/tex]

Therefore,

P(X=6)=0.182819

Step 4:

(c) Probability of at least 6 calls involving a fax message:

We have to add probabilities for x=0 to x=5 and subtract from 1[tex]\begin{aligned}& P(X \geq 6)=1-(P(X < 6)) \\& =1-(P(X \leq 6)-P(X=6)) \\& =1-(0.5611-0.1828) \\& =0.6217\end{aligned}[/tex]

Therefore,[tex]P(X \geq 6)=0.6217[/tex]

Step 5:

(d) Probability of more than 6 calls involving a fax message:

We have to add probabilities for x=0 to x=6 and subtract from 1[tex]\begin{aligned}& P(X \geq 6)=1-(P(X \leq 6)) \\& =1-(0.5611) \\& =0.4389\end{aligned}[/tex]

Therefore,

[tex]P(X \geq 6)=0.4389[/tex]

To learn more about binomial distribution visit:

https://brainly.com/question/14565246

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