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Data Analysis and Probability

Binomial Distribution Function

Getting to Know Jacob Bernoulli

Jacob Bernoulli, also known as James or Jacques, was one of the prominent mathematicians from the Bernoulli family. He was the first pioneer in Leibnizian analysis and supported Leibniz in the calculus debate against Newton.

Jacob Bernoulli is famous for his many contributions to calculus. He was one of the founders of calculus of variations and proposed the first version of the law of large numbers in his book "Ars Conjectandi" published in 1713. One of the important discussions in that book was about binomial experiments.

Combination and Basic Probability

Before diving into binomial distribution, there are some basic formulas we need to master first.

First, the combination formula for selecting objects:

Ckn=(nk)=n!k!(nk)!C_k^n = \binom{n}{k} = \frac{n!}{k!(n-k)!}

Second, factorial (sequential multiplication):

n!=n×(n1)×...×3×2×1n! = n \times (n-1) \times ... \times 3 \times 2 \times 1

And third, the relationship between event probability and its complement:

P(A)+P(Ac)=1P(A) + P(A^c) = 1

where P(A)P(A) is the probability of an event occurring and P(Ac)P(A^c) is the probability of the event not occurring.

Binomial Distribution Concept

Binomial distribution is actually a fairly simple concept if we understand it from the basics. Imagine you're conducting an experiment that only has two possible outcomes: success or failure. For example, like flipping a coin that can only result in heads or tails.

Binomial distribution is used when we perform the same experiment repeatedly under the same constant conditions. Importantly, each experiment is independent (they don't influence each other) - meaning the result of previous experiments doesn't affect subsequent experiments.

Requirements for binomial experiments:

  • There are only two possible outcomes: success or failure

  • The number of experiments is predetermined and fixed

  • Each experiment is mutually independent

  • The probability of success is the same for each experiment

Binomial Distribution Formula

If there's a binomial experiment with probability of success pp and probability of failure q=1pq = 1 - p, then the formula to calculate the probability of getting exactly xx successes in nn independent experiments is:

b(x;n,p)=(nx)pxqnx for x=0,1,2,...,nb(x;n,p) = \binom{n}{x} p^x q^{n-x} \text{ for } x = 0,1,2,...,n

Where:

  • nn = number of experiments performed

  • xx = number of successes we want

  • pp = probability of success in one experiment

  • q=1pq = 1-p = probability of failure in one experiment

So, to use this formula, make sure first that the experiment we're dealing with meets the binomial requirements mentioned earlier.

Coin Experiment

Let's look at a simple example. Suppose we have a fair coin. We call heads as "H" and tails as "T", so the sample space is S={H,T}S = \{H,T\}.

If we consider getting tails as "success", then the probability of success is p=12p = \frac{1}{2}. Automatically, the probability of failure (getting heads) is q=112=12q = 1 - \frac{1}{2} = \frac{1}{2}.

Problem: If we flip this coin 77 times in a row, what's the probability of getting tails exactly 55 times?

Solution:

First, let's identify the parameters:

  • n=7n = 7 (number of flips)
  • x=5x = 5 (number of tails desired)
  • p=12p = \frac{1}{2} (probability of getting tails)
  • q=12q = \frac{1}{2} (probability of getting heads)

Now we use the binomial distribution formula:

b(5;7,12)=(75)(12)5(12)75b(5;7,\frac{1}{2}) = \binom{7}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{7-5}
=(75)(12)5(12)2= \binom{7}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^2
=(75)(12)7= \binom{7}{5} \left(\frac{1}{2}\right)^7

Let's calculate (75)\binom{7}{5} first:

(75)=7!5!(75)!=7!5!2!\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5! \cdot 2!}
=7×6×5!5!×2×1=7×62=422=21= \frac{7 \times 6 \times 5!}{5! \times 2 \times 1} = \frac{7 \times 6}{2} = \frac{42}{2} = 21

So the complete calculation is:

b(5;7,12)=21×127=21×1128=21128b(5;7,\frac{1}{2}) = 21 \times \frac{1}{2^7} = 21 \times \frac{1}{128} = \frac{21}{128}

Therefore, the probability of getting 55 tails in 77 flips is 21128\frac{21}{128} or approximately 16.4%16.4\%.

Practice Problems

  1. In a bag there are 1010 balls that are exactly the same, except for their color: 66 red balls and 44 blue balls. If we draw 55 balls one by one with replacement (each drawn ball is returned), what's the probability of drawing exactly 33 red balls?

  2. An archer has an accuracy rate of 80%80\% for hitting targets. If he shoots 66 times, what's the probability he hits the target exactly 44 times?

  3. There's a multiple choice exam with 44 answer choices for each question. A student guesses all answers randomly for 88 questions. What's the probability he answers exactly 22 questions correctly?

Answer Key

  1. Red Ball Problem Answer

    Step 1: Identify parameters.

    • n=5n = 5 (number of draws)
    • x=3x = 3 (number of red balls desired)
    • p=610=0.6p = \frac{6}{10} = 0.6 (probability of drawing a red ball)
    • q=10.6=0.4q = 1 - 0.6 = 0.4 (probability of drawing a blue ball)

    Step 2: Calculate using binomial distribution formula.

    b(3;5,0.6)=(53)(0.6)3(0.4)2b(3;5,0.6) = \binom{5}{3} (0.6)^3 (0.4)^2

    Let's calculate (53)\binom{5}{3} first:

    (53)=5!3!2!=5×42×1=10\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10

    Then:

    =10×(0.6)3×(0.4)2= 10 \times (0.6)^3 \times (0.4)^2
    =10×0.216×0.16=0.3456= 10 \times 0.216 \times 0.16 = 0.3456

    Therefore, the probability of drawing exactly 33 red balls is 0.34560.3456 or approximately 34.6%34.6\%.

  2. Archer Problem Answer

    Step 1: Identify parameters.

    • n=6n = 6 (number of shots)
    • x=4x = 4 (number of targets desired)
    • p=0.8p = 0.8 (probability of hitting target)
    • q=0.2q = 0.2 (probability of missing)

    Step 2: Calculate using binomial distribution formula.

    b(4;6,0.8)=(64)(0.8)4(0.2)2b(4;6,0.8) = \binom{6}{4} (0.8)^4 (0.2)^2

    Let's calculate (64)\binom{6}{4} first:

    (64)=6!4!2!=6×52×1=15\binom{6}{4} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15

    Then:

    =15×(0.8)4×(0.2)2= 15 \times (0.8)^4 \times (0.2)^2
    =15×0.4096×0.04=0.2458= 15 \times 0.4096 \times 0.04 = 0.2458

    Therefore, the probability of the archer hitting the target exactly 44 times is 0.24580.2458 or approximately 24.6%24.6\%.

  3. Multiple Choice Exam Problem Answer

    Step 1: Identify parameters.

    • n=8n = 8 (number of questions)
    • x=2x = 2 (number of correct answers desired)
    • p=14=0.25p = \frac{1}{4} = 0.25 (probability of answering correctly by guessing)
    • q=34=0.75q = \frac{3}{4} = 0.75 (probability of answering incorrectly)

    Step 2: Calculate using binomial distribution formula.

    b(2;8,0.25)=(82)(0.25)2(0.75)6b(2;8,0.25) = \binom{8}{2} (0.25)^2 (0.75)^6

    Let's calculate (82)\binom{8}{2} first:

    (82)=8!2!6!=8×72×1=28\binom{8}{2} = \frac{8!}{2! \cdot 6!} = \frac{8 \times 7}{2 \times 1} = 28

    Then:

    =28×(0.25)2×(0.75)6= 28 \times (0.25)^2 \times (0.75)^6
    =28×0.0625×0.178=0.3115= 28 \times 0.0625 \times 0.178 = 0.3115

    Therefore, the probability of the student answering exactly 22 questions correctly is 0.31150.3115 or approximately 31.2%31.2\%.