📊 Stats Adventure! Binomial & Expected Value

📖 Concept 1 — Binomial Random Variable

A Binomial Random Variable $X$ counts the number of successes in $n$ independent trials, where each trial has exactly two outcomes (success/failure) and the probability of success $p$ stays constant.

Four conditions (BINS):

B Binary — only two outcomes per trial
I Independent — trials don't affect each other
N Number — fixed number of trials $n$
S Same probability — $p$ is constant

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ $$\mu = np \qquad \sigma^2 = np(1-p) \qquad \sigma = \sqrt{np(1-p)}$$

🌟 Worked Example

A fair coin is flipped 10 times. Let $X$ = number of heads. Then $X \sim B(10,\ 0.5)$.

Mean: $\mu = 10 \times 0.5 = 5$

Standard deviation: $\sigma = \sqrt{10 \times 0.5 \times 0.5} = \sqrt{2.5} \approx 1.58$

$P(X = 3) = \binom{10}{3}(0.5)^3(0.5)^7 = 120 \times 0.125 \times 0.0078125 \approx 0.117$

📖 Concept 2 — Expected Value (Mean)

The Expected Value $E(X)$ is the long-run average outcome — the weighted mean of all possible values.

$$E(X) = \sum x_i \cdot P(X = x_i)$$

For an insurance example: if there's a 1% chance your car is totaled (worth \$10,000):

Expected loss per year = $0.01 \times 10000 + 0.99 \times 0 = \$100$

🌟 Worked Example

Plan A: \$50 premium + \$500 deductible if totaled (1% chance).

Expected cost = \$50 + $0.01 \times 500$ = \$50 + \$5 = \$55

Plan B: \$60 premium + \$0 deductible. Expected cost = \$60

➡️ Plan A has lower expected cost (\$55 < \$60).

Stats Adventure!

📖 Concept 1 — Binomial Random Variable

🌟 Worked Example

📖 Concept 2 — Expected Value (Mean)

🌟 Worked Example

🎯 Choose Your Questions!

🔴 Questions to Review