# Sheldon uses the even numbered cubes below to play a game. 2 , 4 , 6 , 8 and 10. ? Rest of problem is in Details!

## Sheldon uses the even numbered cubes below to play a game. Part A : Sheldon randomly selects a cube 100 times and replaces it. Is it reasonable to expect Sheldon to choose a number that is less than 6 about 50 times? Show work! Part B : Sheldon uses the same process to select a cube 250 times. How many times should he expect to choose the cubes labeled 6, 8, or 10? Justify your reasoning.

Apr 20, 2018

A: Yes it's reasonable

B:$150$ Times

#### Explanation:

$S = \left\{2 , 4 , 6 , 8 , 10\right\}$

color(blue) (Part (A):

Let $C$ be the event of the appearance of a number less than $6$

$C = \left\{2 , 4\right\}$

so $P \left(A\right) = {N}_{C} / {N}_{S}$

$\textcolor{g r e e n}{\text{Where " N_C" is the number of elements of } C = 2}$

$\textcolor{g r e e n}{\text{And " N_S" is the number of elements of } S = 5}$

$P \left(C\right) = \frac{2}{5} = 0.4$

so the probability that $C$ occurs is 40%

so if He selects a cube 100 times then he will get 40 cubes with the number $2 , 4$

but since the question is asking if it's reasonable if He got 50 cubes?

I would say Yes it's reasonable since 50 cubes is near 40

color(blue)(Part(B)

Let $V$ be the event of the appearance of $6 , 8 , 10$

$V = \left\{6 , 8 , 10\right\}$

$P \left(V\right) = {N}_{V} / {N}_{S}$

$P \left(V\right) = \frac{3}{5} = 0.6$

In order to calculate how many times he got a cube labeled $6 , 8 , 10$ when He drew 250 cubes

color(green)("Number of Occuring of " (V)= P(V)xx" Number of draws"

$250 \times 0.6 = 150$ Times

I will put this on double check to make sure $P a r t \left(A\right)$ is correct