Seed Dispersal Complete the Sentence Printable (4th - 6th Grade ... - Free Printable

It seems like the task you're referring to involves a "Sweet Dispenser" scenario, which is likely an educational activity designed to teach concepts related to probability, randomness, or decision-making. Since the image itself isn't visible in this context, I'll reconstruct and solve a typical problem of this nature based on common examples.

Problem Reconstruction:

Imagine a sweet dispenser that randomly dispenses one of three types of candies: chocolate, gum, or candy bar. The dispenser is programmed to select each type with equal probability (i.e., 1/3 for each type). You are tasked with determining the likelihood of certain outcomes after multiple uses of the dispenser.

#### Example Question:
If you use the dispenser three times, what is the probability that you will get at least one chocolate?

---

Solution:

To solve this problem, we need to calculate the probability of getting at least one chocolate in three uses of the dispenser. A straightforward way to approach this is by using the complementary probability method. Instead of directly calculating the probability of getting at least one chocolate, we calculate the probability of the complementary event (getting no chocolates) and subtract it from 1.

#### Step 1: Define the probabilities
- Probability of getting a chocolate in one use: \( P(\text{chocolate}) = \frac{1}{3} \)
- Probability of not getting a chocolate in one use: \( P(\text{not chocolate}) = 1 - \frac{1}{3} = \frac{2}{3} \)

#### Step 2: Calculate the probability of getting no chocolates in three uses
If the dispenser is used three times, the probability of not getting a chocolate in all three uses is the product of the individual probabilities of not getting a chocolate in each use (since the events are independent):

\[
P(\text{no chocolate in 3 uses}) = P(\text{not chocolate}) \times P(\text{not chocolate}) \times P(\text{not chocolate})
\]

\[
P(\text{no chocolate in 3 uses}) = \left( \frac{2}{3} \right) \times \left( \frac{2}{3} \right) \times \left( \frac{2}{3} \right) = \left( \frac{2}{3} \right)^3
\]

\[
P(\text{no chocolate in 3 uses}) = \frac{8}{27}
\]

#### Step 3: Calculate the probability of getting at least one chocolate
The probability of getting at least one chocolate is the complement of the probability of getting no chocolates:

\[
P(\text{at least one chocolate}) = 1 - P(\text{no chocolate in 3 uses})
\]

\[
P(\text{at least one chocolate}) = 1 - \frac{8}{27}
\]

\[
P(\text{at least one chocolate}) = \frac{27}{27} - \frac{8}{27} = \frac{19}{27}
\]

---

Final Answer:

The probability of getting at least one chocolate in three uses of the dispenser is:

\[
\boxed{\frac{19}{27}}
\]

---

Explanation:

1. Complementary Probability: This method simplifies the calculation by focusing on the opposite event (no chocolates) and then subtracting its probability from 1.
2. Independence: Each use of the dispenser is independent, so the probabilities multiply.
3. Equal Likelihood: Since each candy type has an equal chance of being dispensed, the probabilities remain consistent across uses.

This approach can be generalized to similar problems involving random selection and multiple trials. If you have additional details or a different specific question from the worksheet, feel free to provide them!