Question

A cylinder has a height of 8.1 inches and circular base with a diameter of 2.7 inches. The cylinder contains 3 green spheres, each having a diameter of 2.7 inches.
If a point is chosen at random inside the cylinder, then what’s the probability the point belongs outside of the spheres. Show all work necessary.

Answer 1

Answer:

A point is chosen at random inside the cylinder, then probability the point belongs outside of the spheres is 0.25

Step-by-step explanation:

Let's find volume of the cylinder and volume of the three spheres.

Volume of cylinder =$\pi r^{2} h$

Volume of sphere =$\frac{4}{3} \pi r^{3}$

Now, r =2.7

h=8.1

Find volume of each shape.

Volume of cylinder =$\pi (2.7)^{2} (8.1)$

                             =59.049$\pi$

Now, volume of 3 spheres =$3*\frac{4}{3} *\pi *(2.7)^{3}$

                                         =78.732$\pi$

Now, p(E)=$\frac{Outcomes favorable to event E}{Total number of outcomes}$

So, Probability of point is chosen at random inside the cylinder which belongs outside of the spheres  is

                  $\frac{78.732\pi -59.049\pi }{78.732\pi }$

Let's simplify them

             0.25

A point is chosen at random inside the cylinder, then probability the point belongs outside of the spheres is 0.25