Question 11 (Multiple Choice Worth 2 Points)

A bag consists of 5 marbles: 1 yellow, 1 red, 1 clear, 1 green, and 1 blue marble. What is the theoretical probability of choosing 2 marbles from the bag with replacement?

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
& Yellow & Red & Clear & Green & Blue \\
\hline
Yellow & Yellow & Yellow & Yellow & Yellow & Yellow \\
\hline
Red & Red & Red & Red & Red & Red \\
\hline
Clear & Clear & Clear & Clear & Clear & Clear \\
\hline
Green & Green & Green & Green & Green & Green \\
\hline
Blue & Blue & Blue & Blue & Blue & Blue \\
\hline
\end{tabular}
\][/tex]



Answer :

Let's break down the problem step-by-step to understand the theoretical probability of choosing 2 marbles from the bag with replacement.

The bag consists of 5 marbles:
- 1 Yellow
- 1 Red
- 1 Clear
- 1 Green
- 1 Blue

Since the marbles are being replaced, the total number of marbles remains constant at 5 for each draw.

Step 1: Determine the probabilities of selecting each individual marble.

Each marble has an equal chance of being selected. Therefore, the probability of selecting any single marble (Yellow, Red, Clear, Green, or Blue) from the bag is:
[tex]\[ \text{Probability of any marble} = \frac{1 \text{ marble}}{5 \text{ total marbles}} = 0.2 \][/tex]

Step 2: Consider the combination of selecting two marbles.

Since we're selecting with replacement and the selections are independent:

- The probability of selecting a Yellow marble first and a Yellow marble second is:
[tex]\[ P(Yellow \text{ and } Yellow) = P(Yellow) \times P(Yellow) = 0.2 \times 0.2 = 0.04 \][/tex]

- Similarly, the probability of selecting a Yellow marble first and Red marble second is:
[tex]\[ P(Yellow \text{ and } Red) = P(Yellow) \times P(Red) = 0.2 \times 0.2 = 0.04 \][/tex]

Step 3: Generalize for all combinations.

This reasoning applies to all possible combinations of selecting any two marbles from the bag (with replacement), including:
[tex]\[ P(Yellow \text{ and } Yellow), P(Yellow \text{ and } Red), P(Yellow \text{ and } Clear), P(Yellow \text{ and } Green), P(Yellow \text{ and } Blue), \][/tex]
[tex]\[ P(Red \text{ and } Yellow), P(Red \text{ and } Red), P(Red \text{ and } Clear), P(Red \text{ and } Green), P(Red \text{ and } Blue), \][/tex]
[tex]\[ P(Clear \text{ and } Yellow), P(Clear \text{ and } Red), P(Clear \text{ and } Clear), P(Clear \text{ and } Green), P(Clear \text{ and } Blue), \][/tex]
[tex]\[ P(Green \text{ and } Yellow), P(Green \text{ and } Red), P(Green \text{ and } Clear), P(Green \text{ and } Green), P(Green \text{ and } Blue), \][/tex]
[tex]\[ P(Blue \text{ and } Yellow), P(Blue \text{ and } Red), P(Blue \text{ and } Clear), P(Blue \text{ and } Green), P(Blue \text{ and } Blue) \][/tex]

Each of these combinations has a probability of:
[tex]\[ 0.2 \times 0.2 = 0.04 \][/tex]

In conclusion, every specific combination of two marbles has an overall probability of 0.04.

Thus, if you add the probabilities for choosing each specific marble, you get:
[tex]\[ \boxed{'Yellow': 0.2, 'Red': 0.2, 'Clear': 0.2, 'Green': 0.2, 'Blue': 0.2} \][/tex]