Question

Kittens. In a litter of seven kittens, three are female. You pick two kittens at random. a) Create a probability model for the number of male kittens you get. b) What's the expected number of males? c) What's the standard deviation?

Short answer

a) Probabilities: P(X=0)=1/7, P(X=1)=4/7, P(X=2)=2/7. b) Expected males: 1.14. c) Std deviation: 0.64.

Step-by-step solution

Understanding the Problem

We have a total of 7 kittens, out of which 3 are females and 4 are males. We need to pick 2 kittens at random and find the probability of getting 0, 1, or 2 male kittens.

Defining the Random Variable

Let the random variable \( X \) represent the number of male kittens picked. The possible values for \( X \) are 0, 1, or 2, corresponding to picking no male kittens, one male kitten, or two male kittens respectively.

Calculating Probabilities

We need to calculate the probability for each value of \( X \):1. \( P(X=0) \): Probability of picking 0 male kittens. This occurs when both picked kittens are female. There are \( \binom{3}{2} = 3 \) ways to pick 2 females from 3, and there are \( \binom{7}{2} = 21 \) ways to pick any 2 kittens from 7. So, \( P(X=0) = \frac{3}{21} = \frac{1}{7} \).2. \( P(X=1) \): Probability of picking 1 male and 1 female kitten. There are \( \binom{4}{1} \binom{3}{1} = 12 \) ways to pick 1 male and 1 female, so \( P(X=1) = \frac{12}{21} = \frac{4}{7} \).3. \( P(X=2) \): Probability of picking 2 male kittens. There are \( \binom{4}{2} = 6 \) ways to pick 2 males. So, \( P(X=2) = \frac{6}{21} = \frac{2}{7} \).

Calculating the Expected Value

The expected number of male kittens, \( E(X) \), is calculated using the formula \( E(X) = \sum xP(X=x) \).\[ E(X) = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} \ = \frac{0}{7} + \frac{4}{7} + \frac{4}{7} = \frac{8}{7} \approx 1.14 \]

Calculating the Variance

Calculate the variance \( \text{Var}(X) \) using \( \text{Var}(X) = E(X^2) - (E(X))^2 \).First find \( E(X^2) \):\[ E(X^2) = 0^2 \times \frac{1}{7} + 1^2 \times \frac{4}{7} + 2^2 \times \frac{2}{7} \ = \frac{0}{7} + \frac{4}{7} + \frac{8}{7} = \frac{12}{7} \]Then, \( \text{Var}(X) = \frac{12}{7} - \left(\frac{8}{7}\right)^2 = \frac{12}{7} - \frac{64}{49} = \frac{84}{49} - \frac{64}{49} = \frac{20}{49} \).

Calculating the Standard Deviation

The standard deviation \( \sigma(X) \) is the square root of the variance:\( \sigma(X) = \sqrt{\frac{20}{49}} \approx 0.64 \).

Key concepts

Expected Value

The expected value is a fundamental concept in probability that gives us the mean or average of a random variable. It's like predicting what you would expect to get if you repeated an experiment many, many times. In the case of our kittens, we're interested in the expected number of male kittens when we pick two at random from the litter.

To calculate the expected value, we use the formula:

• \[ E(X) = \sum xP(X=x) \]

This formula tells us to sum up the products of each possible value of the random variable (in this case, the number of male kittens, which could be 0, 1, or 2) with its corresponding probability.

For our problem:

• When picking 0 male kittens, the probability is \(\frac{1}{7}\).
• For picking 1 male kitten, it's \(\frac{4}{7}\).
• For picking 2 male kittens, the probability is \(\frac{2}{7}\).

So, the expected value is:

• \[ E(X) = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} = \frac{8}{7} \approx 1.14 \]

This means that, on average, if you keep picking 2 kittens over and over, you'd expect there to be about 1.14 male kittens each time.

Variance

Variance helps us understand how much our random variable values (in our case, the number of male kittens) are spread out from the expected value. Essentially, it provides information on how much variation there is from the average number of male kittens.

We start by finding the expected value of the squares of the number of kittens, noted as \(E(X^2)\):

• The squared values and their probabilities: 0 kittens gives \(0^2 \times \frac{1}{7} = \frac{0}{7}\), 1 kitten gives \(1^2 \times \frac{4}{7} = \frac{4}{7}\), and 2 kittens gives \(2^2 \times \frac{2}{7} = \frac{8}{7}\).

So, the expected value of the squares is:

• \[ E(X^2) = \frac{0}{7} + \frac{4}{7} + \frac{8}{7} = \frac{12}{7} \]

After finding \(E(X^2)\), we compute the variance using:

• \[ \text{Var}(X) = E(X^2) - (E(X))^2 \]

For our kittens:

• \[ \text{Var}(X) = \frac{12}{7} - \left(\frac{8}{7}\right)^2 = \frac{20}{49} \]

This value tells us that the outcomes for the number of male kittens fluctuate around the mean with a spread that is represented by \(\frac{20}{49}\).

Standard Deviation

Standard deviation is a metric derived from the variance. It provides a direct measure of the spread in the same units as the original data, making it a bit more intuitive to interpret. In simpler terms, it helps us understand how much the number of male kittens deviates from what we expect on average.

To determine the standard deviation \(\sigma(X)\), we take the square root of the variance:

• \[ \sigma(X) = \sqrt{\text{Var}(X)} \]

For our problem:

• \[ \sigma(X) = \sqrt{\frac{20}{49}} \approx 0.64 \]

The standard deviation of approximately 0.64 means that the number of male kittens typically deviates from the expected 1.14 by around 0.64 kittens. It's a helpful way to gauge the consistency of our outcomes.