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Random Variables and Expected Value

A random variable is just a number that depends on chance — like the sum of two dice or the number of heads in 3 coin flips. The expected value E(X) is the long-run average: if you repeated the experiment thousands of times, the average result would approach $E(X)$. It's a weighted average where each outcome is weighted by its probability. The variance measures how far outcomes tend to spread from that average. Think of it this way: expected value tells you what to "expect on average," and standard deviation tells you "how surprised you might be." For a linear transformation: $E(aX+b) = aE(X)+b$ and $Var(aX+b) = a^2 Var(X)$.

$$E(X)=\sum x\cdot p(x)$$
$$Var(X)=E(X^2)-[E(X)]^2, \quad \sigma=\sqrt{Var(X)}$$

Expected Value from a Table

Let $X=0,1,2,3$ with probabilities $0.10,0.30,0.40,0.20$. Find $E(X)$.

$$E(X)=0(0.10)+1(0.30)+2(0.40)+3(0.20)=1.70$$

Final answer: 1.70.

Variance of a Discrete Variable

For $X=1,2,3$ with probabilities $0.2,0.5,0.3$, find the variance.

$$E(X)=1(0.2)+2(0.5)+3(0.3)=2.1$$
$$E(X^2)=1(0.2)+4(0.5)+9(0.3)=4.9$$
$$Var(X)=4.9-(2.1)^2=0.49$$

Expected Profit

A project earns P80,000 with probability 0.35, P20,000 with probability 0.45, and loses P30,000 with probability 0.20. Find expected profit.

$$E=80000(0.35)+20000(0.45)-30000(0.20)=31,000$$

Final answer: P31,000.

Lottery Expected Value

A lottery ticket costs P50. The prize is P10,000 with probability 0.002 and P500 with probability 0.010. All other outcomes win nothing. Is the ticket a good buy?

Compute expected winnings (not including ticket cost):

$$E(\text{win})=10000(0.002)+500(0.010)+0(0.988)=20+5=25$$

Expected net = 25 − 50 = −P25. On average, you lose P25 per ticket. Final answer: Not a good buy — expected value is negative.

Standard Deviation of a Random Variable

A random variable X has the distribution: X = 0 (p = 0.3), X = 2 (p = 0.5), X = 4 (p = 0.2). Find the standard deviation.

$$E(X)=0(0.3)+2(0.5)+4(0.2)=0+1+0.8=1.8$$
$$E(X^2)=0(0.3)+4(0.5)+16(0.2)=0+2+3.2=5.2$$
$$Var(X)=5.2-1.8^2=5.2-3.24=1.96, \quad \sigma=\sqrt{1.96}=1.4$$

Final answer: $\sigma = 1.4$.

Linear Transformation of Expected Value

A random variable X has E(X) = 5 and Var(X) = 4. Find E(3X + 2) and Var(3X + 2).

Rules: $E(aX+b)=aE(X)+b$ and $Var(aX+b)=a^2Var(X)$. Adding a constant shifts the average but doesn't change the spread.

$$E(3X+2)=3(5)+2=17$$
$$Var(3X+2)=3^2(4)=36, \quad \sigma=6$$

Final answer: E = 17, Var = 36, σ = 6.

Verifying a Probability Distribution

Determine k so that the following is a valid probability distribution: P(X=1)=0.2, P(X=2)=k, P(X=3)=0.3, P(X=4)=0.1.

All probabilities must sum to 1 and each must be non-negative.

$$0.2+k+0.3+0.1=1 \implies k=0.4$$

Check: k = 0.4 ≥ 0 ✓. Final answer: k = 0.4.

Expected Sum of Two Dice

Two fair dice are rolled. Find the expected value and variance of the sum.

For one die: $E(\text{die})=3.5$ and $Var(\text{die})=\frac{35}{12}$. For independent dice:

$$E(\text{sum})=3.5+3.5=7$$
$$Var(\text{sum})=\frac{35}{12}+\frac{35}{12}=\frac{35}{6}\approx5.83$$

Final answer: E(sum) = 7, Var(sum) ≈ 5.83.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q692

MSTE - Statistics and Probability / Probability / Engr. Janclyde Espinosa (Clidez)

If 2 marbles are removed at random from a bag containing black and white marbles, the chance that they are both white is 1/3. If 3 are removed at random, the chance that all are white is 1/6. How many black balls are there?

  1. 4
  2. 6
  3. 5
  4. 7
Let there be $w$ white and $b$ black marbles. Given:
$\frac{w(w-1)}{(w+b)(w+b-1)}=\frac{1}{3}$
$\frac{w(w-1)(w-2)}{(w+b)(w+b-1)(w+b-2)}=\frac{1}{6}$
Dividing the second equation by the first gives:
$\frac{w-2}{w+b-2}=\frac{1}{2}$, so $b=w-2$. Substitution gives $w=6$, hence $b=4$.
$\boxed{4}$

Question Bank: q722

MSTE - Statistics and Probability / Probability / Engr. Janclyde Espinosa (Clidez)

A flower shop has: 2 tulips, 2 roses, 2 daisies, and 2 lilies If two flowers are sold at random, what is the probability of not picking exactly two tulips?

  1. 27/28
  2. 1/8
  3. 1/7
  4. 7/8
There are 8 flowers total. Probability of exactly two tulips is:
$\frac{\binom{2}{2}}{\binom{8}{2}}=\frac{1}{28}$
Probability of not exactly two tulips:
$1-\frac{1}{28}=\frac{27}{28}$
$\boxed{27/28}$

Question Bank: t2144

MSTE - Statistics and Probability / Probability / Besavilla CE Pre-Board Math & Surveying

In a batch of 45 lamps there are 10 faulty lamps. If one lamp is drawn at random, find the probability of it being satisfactory.

  1. 0.7778
  2. 0.5456
  3. 0.1525
  4. 0.7845
  5. 0.3652
Satisfactory lamps $=45-10=35$.
$P(\text{satisfactory})=\frac{35}{45}$
$P(\text{satisfactory})=0.777\ldots$
$\boxed{0.7778}$

Question Bank: w192

MSTE - Statistics and Probability / Probability / MSTE May 2022

An insurance company charges ₱80,000 annually for a car insurance policy and specifies that the company will pay ₱100,000 for a minor accident and ₱500,000 for a major accident. If the probability of a motorist having a minor accident during the year is 0.2 and of having a major accident is 0.05, how much can the insurance company expect to make, on average, on any policy?

  1. ₱30,000
  2. ₱20,000
  3. ₱25,000
  4. ₱35,000
The expected payout is $$100{,}000(0.2)+500{,}000(0.05)=45{,}000.$$ Therefore, the expected profit is $$80{,}000-45{,}000=₱35{,}000.$$

Question Bank: w255

MSTE - Statistics and Probability / Probability / MSTE April 2025

An industrial engineer has found that when she takes a sample of size 5 of a product, 90% of the time there are no defectives in the sample, 3% of the time there is 1 defective, 2% of the time there are 2 defectives, 2% of the time there are 3 defectives, 2% of the time there are 4 defectives, and 1% of the time there are 5 defectives. What is the probability of taking a sample that has at least 3 defectives in the 5?

  1. 7%
  2. 5%
  3. 9%
  4. 15%
Tabulate the discrete probability distribution:

Defectives: 0, 1, 2, 3, 4, 5
Probability: 0.90, 0.03, 0.02, 0.02, 0.02, 0.01

"At least 3" means 3, 4, or 5 defectives:
$P(\text{at least }3) = 0.02 + 0.02 + 0.01$
$\boxed{P = 0.05 = 5\%}$

Question Bank: w259

MSTE - Statistics and Probability / Statistics / MSTE April 2025

The random variable X, representing the number of chocolate chips in a cookie, has the following probability distribution: $P(X=2)=0.01$, $P(X=3)=0.25$, $P(X=4)=0.4$, $P(X=5)=0.3$, $P(X=6)=0.04$. Find the variance of X.

  1. 0.74
  2. 0.64
  3. 0.36
  4. 0.81
The variance of a discrete random variable is
$\sigma^{2} = \sum x^{2}P(x) - \mu^{2}$, where $\mu = \sum xP(x)$.

Mean:
$\mu = 2(0.01)+3(0.25)+4(0.4)+5(0.3)+6(0.04) = 4.11$

Second moment:
$\sum x^{2}P(x) = 4(0.01)+9(0.25)+16(0.4)+25(0.3)+36(0.04) = 17.63$

$\sigma^{2} = 17.63 - (4.11)^{2}$
$\boxed{\sigma^{2} = 0.7379 \approx 0.74}$

Calculator tip: enter the x values in Stat mode with the probabilities as the FREQ column, then read $x\sigma_n^{2}$.

Question Bank: w311

MSTE - Statistics and Probability / Probability / MSTE April 2025

Mrs. Crandall has owned live stocks for a number of years. She has concluded during the years she has owned the stocks that the probabilities that exactly 1, 2, 3, 4, or 5 of her stocks go up in price during a trading session are, respectively, 0.18, 0.30, 0.20, 0.15, and 0.08. Find the probability that at most two stocks go up during a trading session.

  1. 0.48
  2. 0.57
  3. 0.68
  4. 0.97
The given probabilities cover 1 through 5 stocks. Because all probabilities must sum to 1, first recover $P(0)$.

$P(1) + P(2) + P(3) + P(4) + P(5)$
$= 0.18 + 0.30 + 0.20 + 0.15 + 0.08 = 0.91$
$\therefore P(0) = 1 - 0.91 = 0.09$

"At most two" means 0, 1, or 2:
$P(0 \le 2) = P(0) + P(1) + P(2)$
$P = 0.09 + 0.18 + 0.30$
$\boxed{P = 0.57}$