Question

Batteries. In a group of 10 batteries, 3 are dead. You choose 2 batteries at random. a) Create a probability model for the number of good batteries you get. b) What's the expected number of good ones you get? c) What's the standard deviation?

Short answer

Expected number of good batteries is 1.4; standard deviation is 0.61.

Step-by-step solution

Understanding the Problem

You have 10 batteries in total, out of which 3 are dead. This means that 7 batteries are good. We need to determine the probability distribution for the number of good batteries when choosing 2 at random.

Define the Random Variable

Let the random variable \( X \) represent the number of good batteries chosen out of 2. \( X \) can take on the values 0, 1, or 2.

Calculate Probabilities

To find the probability of each value of \( X \), use combinations.- \( P(X=0) \): Probability of choosing 0 good batteries, meaning both chosen are dead: \( \frac{\binom{3}{2}}{\binom{10}{2}} = \frac{3}{45} \).- \( P(X=1) \): Probability of choosing 1 good and 1 dead battery. The number of ways to choose 1 good battery and 1 dead battery is \( \binom{7}{1} \times \binom{3}{1} \), so: \( \frac{\binom{7}{1} \times \binom{3}{1}}{\binom{10}{2}} = \frac{21}{45} \).- \( P(X=2) \): Probability of choosing 2 good batteries: \( \frac{\binom{7}{2}}{\binom{10}{2}} = \frac{21}{45} \).

Verify Probability Model

Ensure total probabilities add up to 1:\( P(X=0) + P(X=1) + P(X=2) = \frac{3}{45} + \frac{21}{45} + \frac{21}{45} = 1 \). The model is correct.

Calculate Expected Value

The expected number of good batteries, \( E(X) \), is given by\[ E(X) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2) = 0 \times \frac{3}{45} + 1 \times \frac{21}{45} + 2 \times \frac{21}{45} \]\[ E(X) = \frac{21}{45} + \frac{42}{45} = \frac{63}{45}\approx 1.4 \].

Calculate Variance and Standard Deviation

First, find the variance \( \text{Var}(X) \):\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]Calculate \( E(X^2) \):\[ E(X^2) = 0^2 \times P(X=0) + 1^2 \times P(X=1) + 2^2 \times P(X=2) \]\[ E(X^2) = 0 \cdot \frac{3}{45} + 1 \cdot \frac{21}{45} + 4 \cdot \frac{21}{45} \]\[ E(X^2) = \frac{21}{45} + \frac{84}{45} = \frac{105}{45} = \frac{7}{3} \]Now, calculate the variance:\[ \text{Var}(X) = \frac{7}{3} - \left(\frac{63}{45}\right)^2 \approx \frac{7}{3} - 1.96 \approx 0.37 \]Finally, compute the standard deviation: \( \sigma = \sqrt{\text{Var}(X)} \approx \sqrt{0.37} \approx 0.61 \).

Key concepts

Expected Value

The expected value, often denoted as \( E(X) \), represents the average result you might expect from an experiment if you could repeat it an infinite number of times. For our battery problem, where we're interested in the number of good batteries selected, the expected value helps us anticipate the average number we'd get by selecting two randomly. To compute the expected value, multiply each possible outcome by its probability and then add the results:

• If you select 0 good batteries, the contribution to the expected value is \(0 \times P(X=0)\).
• If you get 1 good battery, the contribution is \(1 \times P(X=1)\).
• For 2 good batteries, it's \(2 \times P(X=2)\).

Adding these contributions gives the formula:\[E(X) = 0 \times \frac{3}{45} + 1 \times \frac{21}{45} + 2 \times \frac{21}{45} = \frac{63}{45} \approx 1.4\]Thus, on average, you can expect to pick approximately 1.4 good batteries from the group when you select two at random. This expected value helps to understand the whole experiment in simplistic terms.

Random Variable

A random variable is a mathematical way to quantify the outcomes of a random event. In our battery example, the random variable \( X \) is defined as the number of good batteries chosen out of two. Random variables can take on various possible values, and each of these values is associated with a probability based on how the scenario is set up. Understanding what a random variable can represent helps in creating and using probability models effectively.For our problem:

• \( X = 0 \): when both batteries selected are dead.
• \( X = 1 \): when one battery is good, and the other is dead.
• \( X = 2 \): when both selected batteries are good.

Random variables are crucial for forming probability models and conducting further computations, such as finding expected values or variances in probability distributions.

Variance and Standard Deviation

Variance and Standard Deviation are measures that help us understand the spread or variability of outcomes from their expected value. In simpler terms, they tell us how much the values of a random variable differ from the average.Variance \( \text{Var}(X) \)Variance quantifies the variability of a distribution around its mean (expected value). It's calculated using the formula:\[\text{Var}(X) = E(X^2) - [E(X)]^2\]For our battery example, we found the expected value squared of \( X \) as:\[E(X^2) = 0^2 \times \frac{3}{45} + 1^2 \times \frac{21}{45} + 2^2 \times \frac{21}{45} = \frac{105}{45} = \frac{7}{3}\]Therefore, the variance \( \text{Var}(X) \) turns out to be:\[\text{Var}(X) = \frac{7}{3} - \left(\frac{63}{45}\right)^2 \approx 0.37\]Standard Deviation \( \sigma \)The standard deviation, \( \sigma \), is the square root of the variance and provides a measure of dispersion in the same units as the data, making it easier to interpret:\[\sigma = \sqrt{\text{Var}(X)} \approx \sqrt{0.37} \approx 0.61\]This number tells us that the number of good batteries we choose will typically vary by about 0.61 from the average. Variance and standard deviation are particularly useful when comparing different probability distributions or assessing the risk in uncertain situations.