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Binomial Distribution

Binomial Distribution. Binomial Distribution The conditions for X to have a binomial distribution are There are only 2 possible outcomes: a success or a failure There is a fixed number of trials , n. The probability of success, p, at each individual trial is constant.

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Binomial Distribution

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  1. Binomial Distribution Binomial Distribution The conditions for X to have a binomial distribution are There are only 2 possible outcomes: a success or a failure There is a fixed number of trials, n. The probability of success, p, at each individual trial is constant. Each trial is independent. Knowing the result of one trial will not influence the next trial.

  2. Binomial Distribution The binomial probability formula is P( X = x) = nCrpx (1 – p)n-x Or P( X = x) = nCrpxqn-x if q = 1 – p n = no of trials, r = no. of successes required

  3. Binomial Distribution Example: A police officer checks 5 cars for Warrant of Fitness. The probability of a car not having a WoF is 1/6 Find the probability that 3 of the cars have not got warrants Answer: P = 1/6 so the probability of failure (to meet the criteria) = 5/6 n = 5 (cars to check), r = 3 (cars without WOF’s) P( X = 3) = 5C3 (1/6)3 (5/6)2 = 0.0322

  4. Binomial Distribution Use the Binomial Probability Formula P( X = x) = nCrpx (1 – p)n-x To evaluate these binomial probabilities P(X=3) for n = 5 and p = 0.2 P(X=3) = 5C3 0.23 0.8 2 = 10 x 0.008 x 0.64 = 0.0 512

  5. Binomial Distribution Use the Binomial Probability Formula P( X = x) = nCrpx (1 – p)n-x To evaluate these binomial probabilities 2. P(X=2) for n = 4 and p = 0.75 P(X=2) = 4C2 0.752 0.25 2 = 6 x 0.5625 x 0.0625 = 0.211

  6. Binomial Distribution Use the Binomial Probability Formula P( X = x) = nCrpx (1 – p)n-x To evaluate these binomial probabilities 3. P(X=4) for n = 6 and p = 0.8 P(X=4) = 6C4 0.84 0.22 = 15 x 0.4096 x 0.04 = 0.246

  7. Binomial Distribution 4. A lawyer successfully defends 1/3 of her clients. Find the probability that 5 out of her next 8 clients are found not guilty. n = 8 r = 5 p = 1/3 q = 2/3 P(X=5) = 8C5 (1/3) 5 (2/3) 3 = 0.0683

  8. Binomial Distribution 5. A motel has 15 rooms. The manager know that the probability that a room will be booked on any night is 0.45 What is the probability that exactly 7 of the rooms will be booked on a given night? n = 15 r = 7 p = 0.45 q = 0.55 P(X=7) = 15C7 (0.45) 7 (0.55) 8 = 0.2013

  9. Binomial Distribution In some examples, you can use the formula sheet directly Ie, in the first example, n = 5, r = 3 and p=1/6 Giving the same answer

  10. Binomial Distribution In June, the probability of it raining on any given day is 0.35. What is the probability that it will rain on exactly 3 days in one week? n = 7 r = 3 p = 0.35 P(X = 3) = 0.2679

  11. Binomial Distribution P(X ≥ 4) = 0.1442 + 0.0466+ 0.0084 + 0.0006 = 0.1998 In June, the probability of it raining on any given day is 0.35. What is the probability that it will rain on at least 4 days in one week? n = 7 r ≥ 4 p = 0.35

  12. Binomial Distribution On a TV production line, the probability of a TV having a fault is 0.05. What is the probability that 4 of the next 10 TV’s will be faulty? P(X = 4) = 0.001

  13. Binomial Distribution On a TV production line, the probability of a TV having a fault is 0.05. What is the probability that no more than 3 of the next 10 TV’s will be faulty? P(X ≤ 3) = 0.5987 + 0.3151 + 0.0746 + 0.0105 = 0.9989

  14. Binomial Distribution The probability that a student will get a detention in any given year is 0.25. If he attends high school for 5 years, what is the probability that he will not get a detention in exactly 3 years? Not getting a detention in 3 years is the same as getting a detention in 2 years.

  15. Binomial Distribution If we check that using the formula, n = 5 p = 0.25 so not getting a detention, q = 0.75 P(X=3) = 5C3(0.75) 3 (0.25) 2 = 0.2637

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