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Binomial Distribution Simplified Revision Notes

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Binomial Distribution

Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successes in nn independent Bernoulli trials, where the probability of success in each trial is pp.

It is used when:

  1. Each trial has only two possible outcomes (success or failure).
  2. The probability of success (pp) remains constant for all trials.
  3. The trials are independent of each other.

Formula

The probability of observing exactly kk successes is given by the formula:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Where:

  • (nk)=n!k!(nk)!\binom{n}{k} = \frac{n!}{k!(n-k)!} is the binomial coefficient.
  • nn is the total number of trials.
  • kk is the number of successes.
  • pp is the probability of success.
  • (1p)(1-p) is the probability of failure.

Mean and Variance

The mean and variance of a binomial distribution are:

Mean: μ=n×p\text{Mean: } \mu = n \times p Variance: σ2=n×p×(1p)\text{Variance: } \sigma^2 = n \times p \times (1-p)

Steps to Solve Binomial Distribution Problems

  1. Define the Variables: Identify nn, pp, and kk
  2. Apply the Formula: Use the binomial probability formula to calculate P(X=k)P(X = k)
  3. Simplify the Expression: Compute factorials and powers as needed.
  4. Interpret the Result: Relate the calculated probability to the context of the problem.

Worked Examples

infoNote

Example 1: Defective Items in a Batch

Problem: A factory produces items with a 5% defect rate. If a batch of 20 items is inspected, what is the probability of finding exactly 2 defective items?


Solution:

Step 1: Identify values: n=20,k=2,p=0.05,1p=0.95n = 20, k = 2, p = 0.05, 1-p = 0.95


Step 2: Apply the formula:

P(X=2)=(202)(0.05)2(0.95)18P(X = 2) = \binom{20}{2} (0.05)^2 (0.95)^{18}

Step 3: Compute:

(202)=20!2!×18!=190\binom{20}{2} = \frac{20!}{2! \times 18!} = 190

Step 4: Simplify:

P(X=2)=190×(0.05)2×(0.95)18190×0.0025×0.418P(X = 2) = 190 \times (0.05)^2 \times (0.95)^{18} \approx 190 \times 0.0025 \times 0.418P(X=2)0.198P(X = 2) \approx 0.198

Answer: The probability is approximately 0.198 or 19.8%.


infoNote

Example 2: Flipping Coins

Problem: What is the probability of getting exactly 3 heads in 5 flips of a fair coin?


Solution:

Step 1: Identify values: n=5,k=3,p=0.5,1p=0.5n = 5, k = 3, p = 0.5, 1-p = 0.5


Step 2: Apply the formula:

P(X=3)=(53)(0.5)3(0.5)2P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2

Step 3: Compute:

(53)=5!3!×2!=10\binom{5}{3} = \frac{5!}{3! \times 2!} = 10

Step 4: Simplify:

P(X=3)=10×(0.5)5=10×0.03125=0.3125P(X = 3) = 10 \times (0.5)^5 = 10 \times 0.03125 = 0.3125

Answer: The probability is 0.3125 or 31.25%.


Summary

  • The binomial distribution is used for problems involving nn independent trials with two outcomes (success or failure).
  • The probability formula is:
P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
  • Mean: μ=n×p\mu = n \times p
  • Variance: σ2=n×p×(1p)\sigma^2 = n \times p \times (1-p)
  • Common applications include quality control, surveys, and games of chance.
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