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aleksandrvk [35]
3 years ago
13

As every Girl Scout knows, statistics teachers seriously love Girl Scout Cookies. The number of boxes of GS cookies statistics t

eachers order, like all important decisions made by statistics teachers, is determined by independent rolls of a 4-sided fair die. If a one appears, 6 boxes are ordered; if any other number appears, 2 boxes are ordered.
a) What is the probability that a statistics teacher places an order for 2 boxes of Girl Scout cookies?
b) What is the probability that with two independently chosen statistics teachers will each order 6 boxes each?
c) What is the probability that for two independently chosen statistics teachers the first will order 6 boxes and the second will order 2 boxes?
d) What is the probability that for two independently chosen statistics teachers exactly one will order 6 boxes?
Mathematics
1 answer:
Alexandra [31]3 years ago
4 0

Answer:

a) 0.75

b) 0.0625

c) 0.1875

d) 0.375

Step-by-step explanation:

Let's start by first finding the probability a teacher orders 6 boxes. Since it is a four sided dice, the probability of getting a 1 is going to be 0.25.

Similarly, for any other number on the dice, the teacher orders 2 boxes. We can think of this as NOT rolling a 1. Using our previous answer, this comes out to be: 1 - 0.25 = 0.75

It should be remembered in the following answers that teachers ordering 6 or 2 boxes are not influenced by the results of other teachers. These are statistically independent events.

a) As given above, this probability is 0.75

b) The probability that both teachers order 6 boxes will be their individual probabilities multiplied. This comes out to be: 0.25 x 0.25 = 0.0625

c) Like in part (b), this is both of their probabilities multiplied:

   0.25 x 0.75 = 0.1875

d) This is nearly the same as part (c), but with a little twist. The order matters. There are two scenarios:

  1.   First teacher gets 6 boxes, and the second gets 2.
  2.   First teacher gets 2 boxes, and the second gets 2.

Both of these events have a probability of 0.1875, as shown in (c). To get the total probability of this event happening, we must add these together:

0.1875 + 0.1875 = 0.375

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Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

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A

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