1. The Binomial Distribution

2. In Statistics we often talk about trials. e.g. A seed is sown and the flower is either yellow or not yellow. We mean an experiment, an investigation or the selection of a sample. However, if we are interested in getting a 6, we could say the trial has only 2 outcomes: a 6 or not a 6. There are 6 possible results ( outcomes ): 1, 2, 3, 4, 5 or 6. e.g. We roll a die. e.g. A computer chip is taken off a production line and it either works or it doesn’t. Lots of trials can be thought of as having 2 outcomes.

3. The Binomial Distribution Suppose that we repeat a trial several times and the probability of success doesn’t change from one trial to the next. The 2 possible outcomes of these trials are called success and failure. Probabilty of success = p and probability failure = q. The trials are independent. Suppose also that each result has no effect on the result of the other trials. What can you say about q + p ? ANS: q + p = 1 since no other outcomes are possible. With these conditions all satisfied, we can use the binomial model to estimate the probability of success and to estimate the mean and variance.

4. The Binomial Distribution SUMMARY  The Binomial distribution can be used to model a situation if all of the following conditions are met: A trial has 2 possible outcomes, success and failure. The probability of success in one trial is p and p is constant for all the trials. The trials are independent. The trial is repeated n times.  n and p are called the parameters of the distribution.

5. The Binomial Distribution e.g. We roll a fair die 4 times and we count the number of fours. There are 4 trials There are 2 outcomes to each trial. ( Success is getting a 4 and failure is not getting a 4 ). The trials are independent. This experiment satisfies the conditions for the binomial model. There is a constant probability of success ( getting a 4 ), so for every trial.

6. The Binomial Distribution Ex. For 4 trials throwing a die X = No. of 4`s obtained 4C0 = 1 no. of ways of obtaining no 4`s 4C1 = 4 no. of ways of obtaining one 4 4C2 = 6 no. of ways of obtaining two 4`s 4C3 = 4 no. of ways of obtaining three 4`s 4C4 = 1 no. of ways of obtaining four 4`s heads 4444

7. The Binomial Distribution 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 For 4 trials throwing a die X = No. of 4`s obtained

8. The Binomial Distribution (q + p) 4 =q4q4 q3p1q3p1 p4p4 q2p2q2p2 q1p3q1p3 4C0+4C2+4C1+4C3+4C4 This means get 4 failures This means get 3 failures and 1 success This means get 1 failure and 3 successes (q + p) 4 =q4q4 q3p1q3p1 p4p4 q2p2q2p2 q1p3q1p3 1+ 6+ 4 + 1 Remember these no.s are also Pascals Triange line 4 P = probability of successq = probability of failure (q + p) 4 =q4q4 q3p1q3p1 p4p4 q2p2q2p2 q1p3q1p3 4C0+4C2+4C1+4C3+4C4 As q + p = 1 q4q4 q3p1q3p1 p4p4 q2p2q2p2 q1p3q1p3 4C0+4C2+4C1+4C3+4C4 =1

9. The Binomial Distribution q4q4 q3p1q3p1 p4p4 q2p2q2p2 q1p3q1p3 4C0+4C2+4C1+4C3+4C4 P(X=x)= For 4 trials throwing a die X = No. of 4`s obtained Where x = 0,1,2,3,4 So the P(X=2) = 4C2q 2 p 2 = 6 (  ) 2 (  ) 2 = 0.1157 So the P(X=4) = 4C4q 0 p 4 = 1 (  ) 0 (  ) 4 = 0.00077 So the P(X=3) = 4C3q 1 p 3 = 4 (  ) 1 (  ) 3 = 0.0154

10. The Binomial Distribution q4q4 q3p1q3p1 p4p4 q2p2q2p2 q1p3q1p3 4C0+4C2+4C1+4C3+4C4 P(X=x)= For 4 trials throwing a die X = No. of 4`s obtained Where x = 0,1,2,3,4 So the P(X=1) = 4C1q 3 p 1 = 4 (  ) 3 (  ) 1 = 0.3858 So the P(X=0) = 4C0q 4 p 0 = 1 (  ) 4 (  ) 0 = 0.4823 Sum of probabilities=0.00077+0.0154+0.1157+0.3858+0.4823 =1

11. The Binomial Distribution 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 P(4444) = 4C4(  ) 0 (  ) 4 For 4 trials throwing a die X = No. of 4`s obtained

12. The Binomial Distribution 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 P(4444) = 4C4(  ) 0 (  ) 4 P(444 ) = 4C3(  ) 1 (  ) 3 For 4 trials throwing a dice X = No. of 4`s obtained

13. The Binomial Distribution 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 P(4444) = 4C4(  ) 0 (  ) 4 P(44 ) = 4C2(  ) 2 (  ) 2 P(444 ) = 4C3(  ) 1 (  ) 3 For 4 trials throwing a dice X = No. of 4`s obtained

14. The Binomial Distribution 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 P(4444) = 4C4(  ) 0 (  ) 4 P(44 ) = 4C2(  ) 2 (  ) 2 P(4 ) = 4C1(  ) 3 (  ) 1 P(444 ) = 4C3(  ) 1 (  ) 3 For 4 trials throwing a dice X = No. of 4`s obtained

15. The Binomial Distribution 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 P(4444) = 4C4(  ) 0 (  ) 4 P(44 ) = 4C2(  ) 2 (  ) 2 P(4 ) = 4C1(  ) 3 (  ) 1 P( ) = 4C0(  ) 4 (  ) 0 P(444 ) = 4C3(  ) 1 (  ) 3 For 4 trials throwing a dice X = No. of 4`s obtained

16. The Binomial Distribution So the P(X=2) = 4C2q 2 p 2 = 6 (  ) 2 (  ) 2 = 0.1157 How does this compare with the experimental value Experimental demo N = 10000 trials P(X=2) = 0.12 Close – why not closer? p=  = 0.16666 In expt p = 0.17

17. The Binomial Distribution Setting up a Binomial Distribution A probability distribution gives the probabilities for all possible values of a variable. Can you use the formula we developed for rolling a die 4 times to work for a die thrown 1) 3 timesX = no. of 6`s 2) 6 timesX = no. of 1`s

18. The Binomial Distribution For 3 trials throwing a dice X = No. of 6`s obtained +3C2+3C1+3C3 (q + p) 3 = P(X=x) = nCx q n–x p x P(X=x) = 3Cx (  ) 3–x (  ) x q0p3q0p3 q1p2q1p2 q2p1q2p1 q3p0q3p0 3C0 Notation used in formula book nCx = So P(X=x) =

19. The Binomial Distribution For 6 trials throwing a dice X = No. of 1`s obtained q6p0q6p0 q5p1q5p1 q4p2q4p2 q3p3q3p3 6C0+6C2+6C1+6C3 (q + p) 6 = q2p4q2p4 +6C4q1p5q1p5 +6C5q0p6q0p6 +6C6 P(X=x) = nCx q n–x p x P(X=x) = 6Cx (  ) 6–x (  ) x So P(X=x) = Using the notation in the formula book

20. The Binomial Distribution If X is the random variable “ the number of sixes when a die is rolled 5 times ” then X has a binomial distribution and Tip: For any binomial probability, these numbers... are equal

21. The Binomial Distribution If X is the random variable “ the number of sixes when a die is rolled 5 times ” then X has a binomial distribution and and this... is the sum of these

22. The Binomial Distribution We can find the probabilities of getting 0, 1, 2, 4 and 5 sixes in the same way. Tip: It saves some fiddling on the calculator if you remember that Can you find the probabilities that X = 0 a nd X = 4 and X = 5 ? ( Give the answers correct to 4 d.p. ) ( 4 d.p. ) We can simplify the expression using a calculator: It’s useful to remember that

23. The Binomial Distribution The probability isn’t exactly zero so we need to show the 4 noughts to give the answer correct to 4 d.p. The probabilities are: Since the sum of the probabilities is 1, I added the others and subtracted from 1. Tip: If you have answers listed like this you need not write them out in a table.

24. The Binomial Distribution In general, if X is a random variable with a binomial distribution, then we write where n is the number of trials and p is the probability of success in one trial. The probabilities of 0, 1, 2, 3,... n successes are given by where x = 0, 1, 2, 3,... n and q = 1  p

25. The Binomial Distribution In order to find this probability we have to add 2 results. To be sure of the accuracy of the answer, we must use 4 decimal places in the individual calculations. e.g.1 If find the probability that X equals 0 or 1 giving the answer correct to 3 d.p. Solution: When adding numbers, always use 1 more d.p. than you need in the answer OR store each individual number in your calculator’s memories. If we had used 3 d.p. for the individual probabilities we would have got for the answer, which is incorrect.

26. The Binomial Distribution Solution: e.g.2 If, find (a) (b)(c) (a) Don’t forget that the binomial always has X = 0 as one possibility. (b) < 2 = 0 or 1

27. The Binomial Distribution c) Can you see the quick way of doing this? ANS: Subtract the probabilities that we don’t want from 1. We found this in part (b) Solution: e.g.2 If, find (a) (b)(c)

28. The Binomial Distribution Exercise (a) (b) 1.If find (c) (a) Solution: (b) (c)