Math Magic: Probability Adventures

Exercise 1: Mean and Variance

Part (i)

Part (ii)

Part (iii)

Part (iv)

For the random variable \(X\) with PMF: \(f(x)=\begin{cases} \frac{1}{10} & x=2,5 \\ \frac{1}{5} & x=0,1,3,4 \end{cases}\), find the mean and variance.

First, let's list all possible values of X with their probabilities:

\[ \begin{array}{c|c} x & P(X=x) \\ \hline 0 & \frac{1}{5} \\ 1 & \frac{1}{5} \\ 2 & \frac{1}{10} \\ 3 & \frac{1}{5} \\ 4 & \frac{1}{5} \\ 5 & \frac{1}{10} \\ \end{array} \]

Calculate the mean (expected value) \(E[X]\):

\[ E[X] = \sum x \cdot P(X=x) = 0 \times \frac{1}{5} + 1 \times \frac{1}{5} + 2 \times \frac{1}{10} + 3 \times \frac{1}{5} + 4 \times \frac{1}{5} + 5 \times \frac{1}{10} \] \[ = 0 + \frac{1}{5} + \frac{2}{10} + \frac{3}{5} + \frac{4}{5} + \frac{5}{10} = 2.3 \]

Calculate \(E[X^2]\) for variance:

\[ E[X^2] = \sum x^2 \cdot P(X=x) = 0^2 \times \frac{1}{5} + 1^2 \times \frac{1}{5} + 2^2 \times \frac{1}{10} + 3^2 \times \frac{1}{5} + 4^2 \times \frac{1}{5} + 5^2 \times \frac{1}{10} \] \[ = 0 + \frac{1}{5} + \frac{4}{10} + \frac{9}{5} + \frac{16}{5} + \frac{25}{10} = 8.1 \]

Now calculate variance \(Var(X)\):

\[ Var(X) = E[X^2] - (E[X])^2 = 8.1 - (2.3)^2 = 8.1 - 5.29 = 2.81 \]

Mean (\(E[X]\)) = 2.3
Variance = 2.81

For the random variable \(X\) with PMF: \(f(x)=\begin{cases} \frac{4-x}{6} & x=1,2,3 \end{cases}\), find the mean and variance.

First, calculate probabilities for each x value:

\[ P(X=1) = \frac{4-1}{6} = \frac{3}{6} = \frac{1}{2} \] \[ P(X=2) = \frac{4-2}{6} = \frac{2}{6} = \frac{1}{3} \] \[ P(X=3) = \frac{4-3}{6} = \frac{1}{6} \]

Calculate the mean \(E[X]\):

\[ E[X] = 1 \times \frac{1}{2} + 2 \times \frac{1}{3} + 3 \times \frac{1}{6} = \frac{1}{2} + \frac{2}{3} + \frac{3}{6} = \frac{10}{6} \approx 1.6667 \]

Calculate \(E[X^2]\):

\[ E[X^2] = 1^2 \times \frac{1}{2} + 2^2 \times \frac{1}{3} + 3^2 \times \frac{1}{6} = \frac{1}{2} + \frac{4}{3} + \frac{9}{6} = \frac{20}{6} \approx 3.3333 \]

Calculate variance:

\[ Var(X) = E[X^2] - (E[X])^2 = \frac{20}{6} - \left(\frac{10}{6}\right)^2 = \frac{20}{6} - \frac{100}{36} = \frac{5}{9} \approx 0.5556 \]

Mean (\(E[X]\)) = \(\frac{10}{6} \approx 1.6667\)
Variance = \(\frac{5}{9} \approx 0.5556\)

For the random variable \(X\) with PDF: \(f(x)=\begin{cases} 2(x-1) & 1

First, verify this is a valid PDF by checking it integrates to 1:

\[ \int_{1}^{2} 2(x-1) \, dx = \left[ (x-1)^2 \right]_{1}^{2} = (1)^2 - (0)^2 = 1 \]

Calculate the mean \(E[X]\):

\[ E[X] = \int_{1}^{2} x \cdot 2(x-1) \, dx = 2\int_{1}^{2} (x^2 - x) \, dx \] \[ = 2\left[ \left(\frac{x^3}{3} - \frac{x^2}{2}\right) \right]_{1}^{2} = 2\left[\left(\frac{8}{3} - \frac{4}{2}\right) - \left(\frac{1}{3} - \frac{1}{2}\right)\right] \] \[ = 2\left[\frac{2}{3} - \left(-\frac{1}{6}\right)\right] = 2\left[\frac{5}{6}\right] = \frac{10}{6} \approx 1.6667 \]

Calculate \(E[X^2]\):

\[ E[X^2] = \int_{1}^{2} x^2 \cdot 2(x-1) \, dx = 2\int_{1}^{2} (x^3 - x^2) \, dx \] \[ = 2\left[ \left(\frac{x^4}{4} - \frac{x^3}{3}\right) \right]_{1}^{2} = 2\left[\left(\frac{16}{4} - \frac{8}{3}\right) - \left(\frac{1}{4} - \frac{1}{3}\right)\right] \] \[ = 2\left[\frac{4}{3} - \left(-\frac{1}{12}\right)\right] = 2\left[\frac{17}{12}\right] = \frac{34}{12} \approx 2.8333 \]

Calculate variance:

\[ Var(X) = E[X^2] - (E[X])^2 = \frac{34}{12} - \left(\frac{10}{6}\right)^2 = \frac{34}{12} - \frac{100}{36} = \frac{1}{18} \approx 0.0556 \]

Mean (\(E[X]\)) = \(\frac{10}{6} \approx 1.6667\)
Variance = \(\frac{1}{18} \approx 0.0556\)

This is an example of a triangular distribution, which often appears in project management for task duration estimates!

For the random variable \(X\) with PDF: \(f(x)=\begin{cases} \frac{1}{2}e^{-\frac{x}{2}} & \text{for } x>0 \\ 0 & \text{otherwise} \end{cases}\), find the mean and variance.

First, verify this is a valid PDF:

\[ \int_{0}^{\infty} \frac{1}{2}e^{-x/2} \, dx = \left[ -e^{-x/2} \right]_{0}^{\infty} = 0 - (-1) = 1 \]

Calculate the mean \(E[X]\) (using integration by parts):

\[ E[X] = \int_{0}^{\infty} x \cdot \frac{1}{2}e^{-x/2} \, dx \] Let \(u = x\), \(dv = \frac{1}{2}e^{-x/2}dx\) ⇒ \(du = dx\), \(v = -e^{-x/2}\) \[ = \left[ -xe^{-x/2} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-x/2} \, dx \] \[ = (0 - 0) + \left[ -2e^{-x/2} \right]_{0}^{\infty} = 0 + (0 - (-2)) = 2 \]

Calculate \(E[X^2]\) (using integration by parts twice):

\[ E[X^2] = \int_{0}^{\infty} x^2 \cdot \frac{1}{2}e^{-x/2} \, dx \] First integration by parts: \(u = x^2\), \(dv = \frac{1}{2}e^{-x/2}dx\) \[ = \left[ -x^2e^{-x/2} \right]_{0}^{\infty} + \int_{0}^{\infty} 2xe^{-x/2} \, dx \] \[ = 0 + 2\int_{0}^{\infty} xe^{-x/2} \, dx \] Second integration by parts: \(u = x\), \(dv = e^{-x/2}dx\) \[ = 2\left[ \left[ -2xe^{-x/2} \right]_{0}^{\infty} + \int_{0}^{\infty} 2e^{-x/2} \, dx \right] \] \[ = 2\left[ 0 + \left[ -4e^{-x/2} \right]_{0}^{\infty} \right] = 2[0 - (-4)] = 8 \]

Calculate variance:

\[ Var(X) = E[X^2] - (E[X])^2 = 8 - (2)^2 = 8 - 4 = 4 \]

Mean (\(E[X]\)) = 2
Variance = 4

This is an exponential distribution with λ = 1/2, commonly used to model waiting times between events in a Poisson process!

Exercise 2: Probability Mass Function

Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let \(X\) be the number of red balls drawn. Find the probability mass function and mean for \(X\).

First, determine all possible values of X:

\[ X \in \{0, 1, 2\} \]

Calculate total number of ways to draw 2 balls from 7:

\[ \text{Total combinations} = C(7,2) = \frac{7!}{2!5!} = 21 \]

Calculate probabilities for each case:

\[ P(X=0) = \frac{C(3,2)}{C(7,2)} = \frac{3}{21} = \frac{1}{7} \] \[ P(X=1) = \frac{C(4,1)C(3,1)}{C(7,2)} = \frac{12}{21} = \frac{4}{7} \] \[ P(X=2) = \frac{C(4,2)}{C(7,2)} = \frac{6}{21} = \frac{2}{7} \]

Now, the PMF is:

\[ f(x) = \begin{cases} \frac{1}{7} & \text{for } x=0 \\ \frac{4}{7} & \text{for } x=1 \\ \frac{2}{7} & \text{for } x=2 \\ 0 & \text{otherwise} \end{cases} \]

Calculate the mean \(E[X]\):

\[ E[X] = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} = 0 + \frac{4}{7} + \frac{4}{7} = \frac{8}{7} \approx 1.1429 \]

PMF: \(f(0)=\frac{1}{7}\), \(f(1)=\frac{4}{7}\), \(f(2)=\frac{2}{7}\)
Mean = \(\frac{8}{7} \approx 1.1429\)

Did you know? This is an example of a hypergeometric distribution, which describes scenarios where you sample without replacement from a finite population with two distinct groups!

Exercise 3: Mean and Variance from Expectations

If \(\mu\) and \(\sigma^2\) are the mean and variance of the discrete random variable \(X\), and \(E(X+3)=10\) and \(E[(X+3)^2]=116\), find \(\mu\) and \(\sigma^2\).

First, expand \(E[X+3]\):

\[ E[X+3] = E[X] + 3 = \mu + 3 = 10 \Rightarrow \mu = 10 - 3 = 7 \]

Now expand \(E[(X+3)^2]\):

\[ E[(X+3)^2] = E[X^2 + 6X + 9] = E[X^2] + 6E[X] + 9 = 116 \] \[ \text{But } E[X] = \mu = 7, \text{ so:} \] \[ E[X^2] + 6 \times 7 + 9 = 116 \Rightarrow E[X^2] + 42 + 9 = 116 \Rightarrow E[X^2] = 65 \]

Recall that variance \(\sigma^2 = E[X^2] - (E[X])^2\):

\[ \sigma^2 = 65 - (7)^2 = 65 - 49 = 16 \]

Mean (\(\mu\)) = 7
Variance (\(\sigma^2\)) = 16

This problem demonstrates how linear transformations affect the mean and variance of random variables!

Exercise 4: Coin Toss Experiment

Four fair coins are tossed once. Find the probability mass function, mean and variance for number of heads occurred.

This is a binomial experiment with \(n=4\) trials and \(p=0.5\) (probability of heads):

The PMF is given by:

\[ P(X=k) = C(4,k) \times (0.5)^k \times (0.5)^{4-k} = C(4,k) \times (0.5)^4 \text{ for } k=0,1,2,3,4 \]

Calculate probabilities for each \(k\):

\[ P(0) = \frac{C(4,0)}{16} = \frac{1}{16} \] \[ P(1) = \frac{C(4,1)}{16} = \frac{4}{16} = \frac{1}{4} \] \[ P(2) = \frac{C(4,2)}{16} = \frac{6}{16} = \frac{3}{8} \] \[ P(3) = \frac{C(4,3)}{16} = \frac{4}{16} = \frac{1}{4} \] \[ P(4) = \frac{C(4,4)}{16} = \frac{1}{16} \]

For binomial distribution, we know:

\[ \text{Mean } (\mu) = n \times p = 4 \times 0.5 = 2 \] \[ \text{Variance } (\sigma^2) = n \times p \times (1-p) = 4 \times 0.5 \times 0.5 = 1 \]

PMF: \(P(k) = \frac{C(4,k)}{16}\) for \(k=0,1,2,3,4\)
Mean = 2
Variance = 1

The binomial distribution was first studied by Jacob Bernoulli in the 17th century, which is why it's sometimes called the "Bernoulli distribution" (though technically that refers to the n=1 case)!

Exercise 5: Uniform Distribution

A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let \(X\) denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the pdf of \(X\) is \(f(x)=\begin{cases} \frac{1}{30} & 0

This is a uniform distribution between 0 and 30 minutes.

For a uniform distribution on \([a,b]\), the expected value is:

\[ E[X] = \frac{a + b}{2} = \frac{0 + 30}{2} = 15 \text{ minutes} \]

Interpretation:

On average, the student can expect to wait 15 minutes for the train. Since arrival times are uniformly distributed, sometimes he'll wait less, sometimes more, but the long-term average will be 15 minutes.

Expected waiting time = 15 minutes

The uniform distribution is the maximum entropy probability distribution for a random variable with a fixed interval - meaning it represents maximum uncertainty about the outcome within the given bounds!

Exercise 6: Exponential Distribution

The time to failure in thousands of hours of an electronic equipment used in a manufactured computer has the density function \(f(x)=\begin{cases} 3e^{-3x} & x>0 \\ 0 & \text{elsewhere} \end{cases}\). Find the expected life of this electronic equipment.

This is an exponential distribution with rate parameter \(\lambda = 3\).

For an exponential distribution, the expected value is:

\[ E[X] = \frac{1}{\lambda} = \frac{1}{3} \approx 0.3333 \text{ thousand hours} = 333.33 \text{ hours} \]

Alternatively, by integration:

\[ E[X] = \int_{0}^{\infty} x \cdot 3e^{-3x} \, dx \] Using integration by parts with \(u = x\), \(dv = 3e^{-3x}dx\): \[ = \left[ -xe^{-3x} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-3x} \, dx = 0 + \left[ \left(-\frac{1}{3}\right)e^{-3x} \right]_{0}^{\infty} = \frac{1}{3} \]

Expected life = \(\frac{1}{3}\) thousand hours ≈ 333.33 hours

The exponential distribution is "memoryless" - meaning the probability of failure in the next hour is the same no matter how long the equipment has already been running!

Exercise 7: Gamma-like Distribution

The probability density function of the random variable \(X\) is given by \(f(x)=\begin{cases} 16xe^{-4x} & \text{for } x>0 \\ 0 & \text{for } x\leq 0 \end{cases}\). Find the mean and variance of \(X\).

First, recognize this as a Gamma distribution with shape \(k=2\) and rate \(\theta=4\).

For Gamma(\(k,\theta\)), the mean and variance are:

\[ \text{Mean} = \frac{k}{\theta} = \frac{2}{4} = 0.5 \] \[ \text{Variance} = \frac{k}{\theta^2} = \frac{2}{16} = 0.125 \]

Alternatively, by integration (for mean):

\[ E[X] = \int_{0}^{\infty} x \cdot 16xe^{-4x} \, dx = 16\int_{0}^{\infty} x^2e^{-4x} \, dx \] Using Gamma function: \(\int_{0}^{\infty} x^{n-1}e^{-\lambda x} \, dx = \frac{\Gamma(n)}{\lambda^n}\) \[ \text{Here } n=3, \lambda=4 \Rightarrow \int x^2e^{-4x} \, dx = \frac{\Gamma(3)}{4^3} = \frac{2}{64} = \frac{1}{32} \] \[ \text{So } E[X] = 16 \times \frac{1}{32} = 0.5 \]

For \(E[X^2]\):

\[ E[X^2] = \int_{0}^{\infty} x^2 \cdot 16xe^{-4x} \, dx = 16\int_{0}^{\infty} x^3e^{-4x} \, dx \] \[ n=4, \lambda=4 \Rightarrow \int x^3e^{-4x} \, dx = \frac{\Gamma(4)}{4^4} = \frac{6}{256} = \frac{3}{128} \] \[ \text{So } E[X^2] = 16 \times \frac{3}{128} = \frac{3}{8} = 0.375 \]

Calculate variance:

\[ Var(X) = E[X^2] - (E[X])^2 = 0.375 - (0.5)^2 = 0.375 - 0.25 = 0.125 \]

Mean = 0.5
Variance = 0.125

This distribution is a special case of the Erlang distribution, used to model waiting times in queueing systems!

Exercise 8: Expected Profit

A lottery with 600 tickets gives one prize of ₹200, four prizes of ₹100, and six prizes of ₹50. If the ticket costs ₹2, find the expected profit amount of a ticket.

First, calculate probabilities of each prize:

\[ P(\text{₹200}) = \frac{1}{600} \] \[ P(\text{₹100}) = \frac{4}{600} = \frac{1}{150} \] \[ P(\text{₹50}) = \frac{6}{600} = \frac{1}{100} \] \[ P(\text{₹0}) = \frac{600-1-4-6}{600} = \frac{589}{600} \]

Calculate expected winnings \(E[W]\):

\[ E[W] = 200 \times \frac{1}{600} + 100 \times \frac{1}{150} + 50 \times \frac{1}{100} + 0 \times \frac{589}{600} \] \[ = \frac{200}{600} + \frac{100}{150} + \frac{50}{100} + 0 \approx 0.3333 + 0.6667 + 0.50 = ₹1.50 \]

Calculate expected profit (winnings - cost):

\[ \text{Expected profit} = E[W] - \text{ticket cost} = ₹1.50 - ₹2.00 = -₹0.50 \]

Expected profit per ticket = -₹0.50 (a loss of 50 paise on average)

This negative expected value means on average, players lose money - which is how lotteries make profit! The difference between the ticket price and expected winnings is called the "house edge."