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Statistics and Probability - Binomial distribution

Grade 11IB_AA

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

🔑Concepts

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A discrete random variable XX follows a Binomial distribution if it represents the number of successes in nn independent trials, where each trial has the same probability of success pp. This is denoted as X∼B(n,p)X \sim B(n, p).

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

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The probability of failure is defined as q=1−pq = 1 - p.

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The Binomial Coefficient (nr)\binom{n}{r} represents the number of ways to choose rr successes from nn trials.

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The Expected Value E(X)E(X) is the mean of the distribution, representing the average number of successes over many repetitions.

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For calculations in IB AA, the Graphic Display Calculator (GDC) is typically used: binompdf(n, p, r) for P(X=r)P(X = r) and binomcdf(n, p, r) 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 '4' is rolled. Find P(X=3)P(X = 3).

Solution:

Here, n=10n = 10, p=16p = \frac{1}{6}, and r=3r = 3. Using the formula: P(X=3)=(103)(16)3(56)10−3P(X = 3) = \binom{10}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{10-3} P(X=3)=120×1216×78125279936≈0.155P(X = 3) = 120 \times \frac{1}{216} \times \frac{78125}{279936} \approx 0.155

Explanation:

We identify the distribution as X∼B(10,16)X \sim B(10, \frac{1}{6}) and apply the probability mass function for r=3r = 3.

Problem 2:

In a large batch of light bulbs, 5%5\% are defective. A sample of 8080 bulbs is chosen at random. Find the expected number of defective bulbs and the standard deviation.

Solution:

n=80n = 80, p=0.05p = 0.05. E(X)=np=80×0.05=4E(X) = np = 80 \times 0.05 = 4 Var(X)=np(1−p)=80×0.05×0.95=3.8Var(X) = np(1-p) = 80 \times 0.05 \times 0.95 = 3.8 σ=3.8≈1.95\sigma = \sqrt{3.8} \approx 1.95

Explanation:

The expected value is the mean (npnp), and the standard deviation is the square root of the variance (npqnpq).

Problem 3:

A student takes a multiple-choice test with 2020 questions. Each question has 44 options, only one of which is correct. If the student guesses every answer, find the probability that they get at least 88 questions correct.

Solution:

X∼B(20,0.25)X \sim B(20, 0.25). We need to find P(X≥8)P(X \ge 8). P(X≥8)=1−P(X≤7)P(X \ge 8) = 1 - P(X \le 7) Using a GDC (binomcdf): P(X≤7)≈0.89818...P(X \le 7) \approx 0.89818... P(X≥8)=1−0.89818...≈0.102P(X \ge 8) = 1 - 0.89818... \approx 0.102

Explanation:

For 'at least' problems, we use the complement rule: P(X≥r)=1−P(X≤r−1)P(X \ge r) = 1 - P(X \le r-1) because the GDC cumulative function calculates P(X≤x)P(X \le x).