d. Find the probability that a customer rents at most two DVDs. (Give your answer as a decimal) Answer: 0.80

Question

UpStudy AI · Accepted Answer

To determine whether the probability that a customer rents **at most two DVDs** is **0.80**, we need to consider the probability distribution governing the number of DVDs rented by a customer.

Assuming you have a probability distribution (such as a binomial distribution, Poisson distribution, or empirical probabilities from data), the probability of renting at most two DVDs is calculated by summing the probabilities of renting 0, 1, or 2 DVDs:

$$
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
$$

**Here's a general approach to verify the probability:**

1. **Identify the Distribution:**
   - **Binomial Distribution:** If there's a fixed number of trials (e.g., customers renting DVDs), each with the same probability of renting a DVD.
   - **Poisson Distribution:** Often used for counting the number of events (like rentals) in a fixed interval of time or space.
   - **Empirical Data:** Based on observed frequencies from real data.

2. **Calculate Individual Probabilities:**
   - **For Binomial:** Use the formula 
     $$
     P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
     $$
     where $$ n $$ is the number of trials, $$ p $$ is the probability of renting a single DVD, and $$ k $$ is the number of DVDs rented.
   - **For Poisson:** Use the formula 
     $$
     P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
     $$
     where $$ \lambda $$ is the average rate (mean number of rentals).

3. **Sum the Probabilities:**
   - Add up the probabilities for 0, 1, and 2 DVDs to get $$ P(X \leq 2) $$.

**Example Using Binomial Distribution:**

Suppose the probability of a customer renting a single DVD is $$ p = 0.3 $$, and each customer has the opportunity to rent up to 3 DVDs ($$ n = 3 $$). Then:

$$
\begin{align*}
P(X = 0) &= \binom{3}{0} (0.3)^0 (0.7)^3 = 1 \times 1 \times 0.343 = 0.343 \
P(X = 1) &= \binom{3}{1} (0.3)^1 (0.7)^2 = 3 \times 0.3 \times 0.49 = 0.441 \
P(X = 2) &= \binom{3}{2} (0.3)^2 (0.7)^1 = 3 \times 0.09 \times 0.7 = 0.189 \
\end{align*}
$$

$$
P(X \leq 2) = 0.343 + 0.441 + 0.189 = 0.973
$$

In this example, the probability of renting **at most two DVDs** is **0.973**, not **0.80**.

**Possible Reasons for Discrepancy:**

1. **Different Distribution Parameters:** Ensure you're using the correct parameters for the distribution (e.g., $$ n $$, $$ p $$, or $$ \lambda $$).
2. **Truncated Distribution:** If there's a maximum limit to the number of DVDs that can be rented, it affects the probabilities.
3. **Empirical Data:** If using real data, ensure that the frequencies are correctly calculated.

**Conclusion:**

Without the specific details of the probability distribution or parameters you're working with, it's challenging to confirm whether **0.80** is the correct probability for renting at most two DVDs. Please provide additional information about the distribution or the context of the problem, and I can offer a more precise verification or guidance.
The probability that a customer rents at most two DVDs is 0.80.

d. Find the probability that a customer rents at most two DVDs. (Give your answer as a decimal) Answer: 0.80

Solución de tutoría real

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Probability Questions