Respuesta :

JeanaShupp

Answer: 0.973

Step-by-step explanation:

The probability mass function of geometric distribution is

[tex]f(x)=(1-p)^{x-1}p,\ \ \ x=1,2,3,...[/tex], p= probability of success on a trial.

As per given, we have

p=0.7

The probability that we'll have at most 3 orders [tex]P(X\leq3)=P(x=1)+P(x=2)+P(x=3)[/tex]

[tex]=(1-0.7)^{1-1}(0.7)+(1-0.7)^{2-1}(0.7)+(1-0.7)^{3-1}(0.7)\\\\=0.7+0.21+0.063\\\\=0.973[/tex]

hence, the required probability = 0.973

Cricetus

The probability that we'll have at most 3 orders will be "0.973".

According to the question,

→ [tex]x \sim Geometric (p)[/tex]

here,

p = 0.7

then,

→ [tex]P(X=x) = (1-p)^{x-1}\times p[/tex]

By putting the given values, we get

                   [tex]= (1-0.7)^{x-1}\times 0.7[/tex]

hence,

The probability that we'll have at most 3 orders will be:

→ [tex]P(X \leq 3)=P(X=1)+P(X=2)+P(X=3)[/tex]

                   [tex]=0.7+0.21+0.063[/tex]

                   [tex]=0.973[/tex]

Thus the above response is correct.

Learn more about probability here:

https://brainly.com/question/6981569

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