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Statistics and Probability - Binomial distribution

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Binomial distribution models the number of successes in a fixed number of independent trials. It is denoted as X∼B(n,p)X \sim B(n, p), where nn is the number of trials and pp is the probability of success.

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The conditions for a Binomial distribution (often remembered as BINS) are: Binary outcomes (success or failure), Independent trials, fixed Number of trials (nn), and constant probability of Success (pp).

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The discrete random variable XX can take any integer value rr such that 0≤r≤n0 \le r \le n.

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The complement probability (failure) is often denoted as qq, where q=1−pq = 1 - p.

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IB AI students are expected to use a Graphic Display Calculator (GDC) for most calculations using functions like 'Binomial PDF' (for P(X=r)P(X = r)) and 'Binomial CDF' (for P(X≤r)P(X \le r)).

📐Formulae

P(X=r)=(nr)pr(1−p)n−rP(X = r) = \binom{n}{r} p^r (1-p)^{n-r}

(nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

E(X)=npE(X) = np

Var(X)=np(1−p)Var(X) = np(1-p)

σ=np(1−p)\sigma = \sqrt{np(1-p)}

💡Examples

Problem 1:

A fair six-sided die is rolled 1010 times. Let XX be the number of times a 44 is rolled. Find the probability that a 44 is rolled exactly 33 times.

Solution:

P(X=3)=(103)(16)3(56)7≈0.155P(X = 3) = \binom{10}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^7 \approx 0.155

Explanation:

Identify the parameters: n=10n = 10, p=16p = \frac{1}{6}, and r=3r = 3. Use the Binomial Probability Density Function (binompdf) on the GDC with these values.

Problem 2:

In a large shipment of light bulbs, 5%5\% are known to be defective. A random sample of 2020 bulbs is tested. Find the probability that at most 22 bulbs are defective.

Solution:

P(X≤2)=P(X=0)+P(X=1)+P(X=2)≈0.925P(X \le 2) = P(X=0) + P(X=1) + P(X=2) \approx 0.925

Explanation:

This is a cumulative probability problem where n=20n = 20 and p=0.05p = 0.05. Use the Binomial Cumulative Distribution Function (binomcdf) on the GDC with lower bound 00 and upper bound 22.

Problem 3:

A student takes a multiple-choice test with 5050 questions. Each question has 44 options, and only one is correct. If the student guesses every answer, calculate the expected number of correct answers and the standard deviation.

Solution:

E(X)=50×0.25=12.5E(X) = 50 \times 0.25 = 12.5 σ=50×0.25×0.75=9.375≈3.06\sigma = \sqrt{50 \times 0.25 \times 0.75} = \sqrt{9.375} \approx 3.06

Explanation:

The number of trials is n=50n = 50 and the probability of success is p=14=0.25p = \frac{1}{4} = 0.25. The mean (expectation) is npnp and the standard deviation is np(1−p)\sqrt{np(1-p)}.