34% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is (a) exactly two, (b) more than two, and (c) between two and five, inclusive. (a) P(2)= (Round to the nearest thousandth as needed.)

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### Probability in Sales Tax Holiday Purchase Likelihood

**Scenario:**
34% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is:
(a) exactly two,
(b) more than two,
(c) between two and five, inclusive.

**Calculations:**

(a) P(2) = __ (Round to the nearest thousandth as needed).

**Instructions for Students:**

To solve the above problem, you will need to use binomial probability formulas. Here is a quick reminder of the steps involved:

1. **Understanding the Binomial Distribution:**
   - **Number of trials (n):** This represents the number of adults selected, which is 10.
   - **Probability of success (p):** The probability that an adult is more likely to purchase during a sales tax holiday, which is 0.34.
   - **Number of successes (k):** The specific number of adults we are interested in.

2. **Calculating Exact Probability (a):**
   Use the binomial probability formula:
   \[
   P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
   \]
   Where \(\binom{n}{k}\) is the binomial coefficient.

   In this scenario for exactly 2 adults (k = 2):
   \[
   P(2) = \binom{10}{2} (0.34)^2 (0.66)^8
   \]

3. **Calculating Probability of More Than Two (b):**
   This can be calculated by finding the sum of probabilities from 3 to 10:
   \[
   P(X > 2) = \sum_{k=3}^{10} \binom{10}{k} (0.34)^k (0.66)^{10-k}
   \]

4. **Calculating Probability Between Two and Five (c):**
   Here we calculate the sum of probabilities from 2 to 5:
   \[
   P(2 \leq X \leq 5) = \sum_{k=2}^{5} \binom{10}{k} (0.34)^k (0.66)^{10-k
Transcribed Image Text:### Probability in Sales Tax Holiday Purchase Likelihood **Scenario:** 34% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is: (a) exactly two, (b) more than two, (c) between two and five, inclusive. **Calculations:** (a) P(2) = __ (Round to the nearest thousandth as needed). **Instructions for Students:** To solve the above problem, you will need to use binomial probability formulas. Here is a quick reminder of the steps involved: 1. **Understanding the Binomial Distribution:** - **Number of trials (n):** This represents the number of adults selected, which is 10. - **Probability of success (p):** The probability that an adult is more likely to purchase during a sales tax holiday, which is 0.34. - **Number of successes (k):** The specific number of adults we are interested in. 2. **Calculating Exact Probability (a):** Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where \(\binom{n}{k}\) is the binomial coefficient. In this scenario for exactly 2 adults (k = 2): \[ P(2) = \binom{10}{2} (0.34)^2 (0.66)^8 \] 3. **Calculating Probability of More Than Two (b):** This can be calculated by finding the sum of probabilities from 3 to 10: \[ P(X > 2) = \sum_{k=3}^{10} \binom{10}{k} (0.34)^k (0.66)^{10-k} \] 4. **Calculating Probability Between Two and Five (c):** Here we calculate the sum of probabilities from 2 to 5: \[ P(2 \leq X \leq 5) = \sum_{k=2}^{5} \binom{10}{k} (0.34)^k (0.66)^{10-k
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