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Statistics and Probability - Discrete Probability Distributions (Binomial, Poisson)

Grade 12IB

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

🔑Concepts

Discrete Random Variables (DRV): A DRV XX takes on a countable set of distinct values. The probability distribution is defined by a function P(X=x)P(X=x) such that the sum of all probabilities equals 11. Visually, this is represented by a probability mass function (PMF) where vertical bars at each xx value show the probability, and the total height of all bars is exactly 11.

Expectation and Variance: The expected value E(X)E(X) represents the theoretical mean or the 'balance point' of the probability distribution's histogram. The variance Var(X)Var(X) and standard deviation σ\sigma measure the dispersion or 'spread' of the distribution around the mean; a larger variance results in a wider, flatter distribution visual.

Binomial Distribution Characteristics: A distribution is Binomial if it satisfies the BINS criteria: Binary outcomes (success/failure), Independent trials, fixed Number of trials nn, and Same probability of success pp. Visually, if p=0.5p=0.5, the distribution is perfectly symmetrical; if p<0.5p<0.5, it is right-skewed; and if p>0.5p>0.5, it is left-skewed.

Binomial Parameters and Notation: Denoted as XB(n,p)X \sim B(n, p). The mean is located at npnp and the variance is np(1p)np(1-p). When nn is large and pp is close to 0.50.5, the discrete bar chart begins to approximate the bell-shaped curve of a normal distribution.

Poisson Distribution Characteristics: Models the number of events occurring in a fixed interval of time or space. It requires that events occur at a constant average rate λ\lambda, are independent, and do not occur simultaneously. Visually, for small λ\lambda, the distribution is highly right-skewed with the highest probability at x=0x=0 or x=1x=1, but as λ\lambda increases, the peak moves to the right and the distribution becomes more symmetrical.

Poisson Parameters and Notation: Denoted as XPo(λ)X \sim Po(\lambda). A unique property of the Poisson distribution is that the mean and the variance are equal (E(X)=Var(X)=λE(X) = Var(X) = \lambda). This means the 'center' and the 'spread' of the distribution are linked to the same rate parameter.

Cumulative Probabilities: In discrete distributions, P(Xk)P(X \le k) is the sum of probabilities for all outcomes from 00 up to kk. Graphically, this corresponds to the cumulative area of the bars to the left of and including the value kk. It is important to distinguish between 'at most' (P(Xk)P(X \le k)), 'less than' (P(X<k)P(X < k)), and 'at least' (P(Xk)P(X \ge k)).

Poisson Approximation to the Binomial: When nn is very large (n>50n > 50) and pp is very small (p<0.1p < 0.1), the Binomial distribution B(n,p)B(n, p) can be approximated by a Poisson distribution Po(λ)Po(\lambda) where λ=np\lambda = np. This is visually useful as it simplifies calculations for rare events in large populations.

📐Formulae

P(X=x)=1\sum P(X=x) = 1

E(X)=xP(X=x)E(X) = \sum x \cdot P(X=x)

Var(X)=E(X2)[E(X)]2Var(X) = E(X^2) - [E(X)]^2

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

E(X)=npE(X) = np (Binomial)

Var(X)=np(1p)Var(X) = np(1-p) (Binomial)

P(X=x)=eλλxx!P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}

E(X)=Var(X)=λE(X) = Var(X) = \lambda (Poisson)

💡Examples

Problem 1:

A biased coin has a probability of 0.60.6 of landing on heads. If the coin is tossed 1010 times, find the probability of getting exactly 77 heads.

Solution:

Let XX be the number of heads, where XB(10,0.6)X \sim B(10, 0.6). We want to find P(X=7)P(X=7). Using the Binomial formula: P(X=7)=(107)(0.6)7(10.6)107P(X=7) = \binom{10}{7} (0.6)^7 (1-0.6)^{10-7} P(X=7)=120(0.6)7(0.4)3P(X=7) = 120 \cdot (0.6)^7 \cdot (0.4)^3 P(X=7)1200.027990.0640.215P(X=7) \approx 120 \cdot 0.02799 \cdot 0.064 \approx 0.215

Explanation:

We identify this as a Binomial problem because there is a fixed number of trials (n=10n=10), two outcomes (heads/tails), and a constant probability (p=0.6p=0.6). We apply the probability mass function for r=7r=7.

Problem 2:

On average, a bakery receives 33 customers every 1515 minutes. Find the probability that the bakery receives exactly 55 customers in a 3030-minute period.

Solution:

First, adjust the rate λ\lambda for the new time interval. If there are 33 customers in 1515 minutes, then in 3030 minutes (which is double the time), the average rate is λ=32=6\lambda = 3 \cdot 2 = 6. Let YY be the number of customers in 3030 minutes, so YPo(6)Y \sim Po(6). We want P(Y=5)P(Y=5): P(Y=5)=e6655!P(Y=5) = \frac{e^{-6} 6^5}{5!} P(Y=5)=0.00247877761200.161P(Y=5) = \frac{0.002478 \cdot 7776}{120} \approx 0.161

Explanation:

This is a Poisson distribution problem because it involves counting independent events in a continuous time interval. The key step is scaling the rate λ\lambda from the 1515-minute interval to the 3030-minute interval before applying the formula.