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Siksha Sarovar is a free e-learning platform for coding courses, BCA university notes and competitive exam preparation. Optional Google sign-in saves your learning progress across devices.

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Discrete Probability Distributions

Lesson 4 of 11 in the free Probability and Statistics- BCA-DS-23-204 notes on Siksha Sarovar, written by Rohit Jangra.

4. Discrete Probability Distributions

1. Binomial Distribution

  • Use: Fixed number of trials ($n$), two outcomes (Success/Failure), independent trials.
  • Example (Quality Control):
  • A factory produces bulbs. 10% are defective. You pick 5 bulbs ($n=5$).
  • Question: What is the probability exactly 2 are defective ($k=2$)?
  • $p = 0.1$ (prob of defect), $1-p = 0.9$.
  • Calculation:
  • $$ P(X=2) = \\binom{5}{2} (0.1)^2 (0.9)^3 $$ $$ P(X=2) = 10 \\times 0.01 \\times 0.729 = 0.0729 $$ (There is a 7.29% chance of finding exactly 2 defective bulbs).

2. Geometric Distribution

  • Use: Repeating trials until the first success occurs.
  • Example (Basketball):
  • A player makes a free throw 20% of the time ($p=0.2$).
  • Question: What is the probability he makes his first shot on the 3rd attempt?
  • This means: Miss, Miss, Make.
  • Calculation:
  • $$ P(X=3) = (1-p)^{3-1} \\times p $$ $$ P(X=3) = (0.8)^2 \\times 0.2 = 0.64 \\times 0.2 = 0.128 $$

3. Poisson Distribution

  • Use: Counting events in a fixed interval (time/area) given an average rate (\\lambda).
  • Example (Call Center):
  • A call center receives an average of 4 calls per minute (\\lambda = 4).
  • Question: What is the probability of receiving exactly 6 calls in a minute?
  • Calculation:
  • $$ P(X=6) = \frac{e^{-4} \cdot 4^6}{6!} $$ $$ P(X=6) = \frac{0.0183 \cdot 4096}{720} \approx 0.104 $$ (There is a 10.4% chance of getting exactly 6 calls).