A bag contains 3 blue marbles, 3 red marbles, and 4 yellow marbles. The probability of drawing a blue marble out of the
bag is or . How many of what color of marbles must be added to the bag so that the probability of a blue marble
being drawn at random from the bag is ?
A. There must be red marbles added to the bag.
B. There must be blue marbles added to the bag.
C. There must be yellow marbles added to the bag.

Ask by Parsons Vaughn. in the United States

Jan 24,2025

Question

A bag contains 3 blue marbles, 3 red marbles, and 4 yellow marbles. The probability of drawing a blue marble out of the
bag is or . How many of what color of marbles must be added to the bag so that the probability of a blue marble
being drawn at random from the bag is ?
A. There must be red marbles added to the bag.
B. There must be blue marbles added to the bag.
C. There must be yellow marbles added to the bag.

Ask by Parsons Vaughn. in the United States

Jan 24,2025

UpStudy AI · Accepted Answer

To achieve a $$75\%$$ probability of drawing a blue marble from the bag, let's analyze each scenario individually.

### **Initial Setup:**
- **Blue marbles:** 3
- **Red marbles:** 3
- **Yellow marbles:** 4
- **Total marbles:** $$3 + 3 + 4 = 10$$
- **Initial probability of drawing a blue marble:** $$\frac{3}{10} = 30\%$$

### **Objective:**
Increase the probability of drawing a blue marble to $$75\%$$.

### **A. Adding Red Marbles**

If we add $$x$$ red marbles:

- **New red marbles:** $$3 + x$$
- **Total marbles:** $$10 + x$$
- **Probability of blue:** $$\frac{3}{10 + x} = 0.75$$

**Solving for $$x$$:**
$$
\frac{3}{10 + x} = 0.75 \
3 = 0.75(10 + x) \
3 = 7.5 + 0.75x \
3 - 7.5 = 0.75x \
-4.5 = 0.75x \
x = -6
$$

**Conclusion:** 
It's **impossible** to achieve a $$75\%$$ probability of drawing a blue marble by adding red marbles alone.

### **B. Adding Blue Marbles**

If we add $$y$$ blue marbles:

- **New blue marbles:** $$3 + y$$
- **Total marbles:** $$10 + y$$
- **Probability of blue:** $$\frac{3 + y}{10 + y} = 0.75$$

**Solving for $$y$$:**
$$
\frac{3 + y}{10 + y} = 0.75 \
3 + y = 0.75(10 + y) \
3 + y = 7.5 + 0.75y \
y - 0.75y = 7.5 - 3 \
0.25y = 4.5 \
y = 18
$$

**Conclusion:** 
You need to **add 18 blue marbles** to achieve a $$75\%$$ probability of drawing a blue marble.

### **C. Adding Yellow Marbles**

If we add $$z$$ yellow marbles:

- **New yellow marbles:** $$4 + z$$
- **Total marbles:** $$10 + z$$
- **Probability of blue:** $$\frac{3}{10 + z} = 0.75$$

**Solving for $$z$$:**
$$
\frac{3}{10 + z} = 0.75 \
3 = 0.75(10 + z) \
3 = 7.5 + 0.75z \
3 - 7.5 = 0.75z \
-4.5 = 0.75z \
z = -6
$$

**Conclusion:** 
It's **impossible** to achieve a $$75\%$$ probability of drawing a blue marble by adding yellow marbles alone.

### **Summary:**
- **A. Red Marbles Added:** **Impossible**
- **B. Blue Marbles Added:** **18**
- **C. Yellow Marbles Added:** **Impossible**

**Answer:**

**B.** To achieve a 75 % probability, 18 blue marbles must be added to the bag.
To achieve a 75% probability of drawing a blue marble, you must add 18 blue marbles to the bag.

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