2. If the probability that James is late home from work on any day is 0.4, what is the probability that he is late home three times in a five-day working week? L L L L’ L’= 0.4 x 0.4 x 0.4 x 0.6 x 0.6 = 0.02304 How many ways could James be late for 3 days out of 5? 1 11 1 21 13 31 14 641 15 10 5 1 1 5 5 1 5L 4L3L2L1L0L P(3 Lates in 5 days) =10 x 0.02304 = 0.2304

3. If there are a large number of stages (or trials), it can be difficult to rely on Pascal’s triangle This example is one of a particular type of problem where there are only two possible outcomes – a BINOMIAL problem. The number of paths giving r occurrences out of n cases is: n C r =

4. The BINOMIAL PROBABILITY DISTRIBUTION is defined as: n stands for the number of trials p stands for the probability of ‘success’ The particular model can be summarised as: The binomial distribution is a DISCRETE distribution.

5. If 25 dice are thrown, find the probability that three sixes are obtained. n = 25

6. You can use the binomial distribution for any situation where you want to count the number of times a particular outcome is observed out of a fixed number of cases – provided certain conditions are satisfied…  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant

7. You can use standard tables to answer some binomial problems… What is (a) P(X ≤ 3)(b) P(X = 3)(c) P(X ≥ 3) (a) P(X ≤ 3) = 0.9460 (b) P(X = 3) = P(X ≤ 3) – P(X ≤ 2) = 0.9460 – 0.7648 = 0.1812 (c) P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.7648 = 0.2352

8. (a) P(X ≤ 17) (a) P(Y ≥ 13) = 1 – P(Y ≤ 12) = 1 – 0.9784 = 0.0216 (b) P(X > 24) (b) P(Y ≤ 5) = 0.2026

9. On 40% if the days that Sean travels to work he finds he has to stop at a particular set of traffic lights. Find the probability that he has to stop at these lights no more that five times during a month in which he works 20 days.

10. MEAN AND VARIANCE OF THE BINOMIAL DISTRIBUTION E(X) = npVar(X) = np(1 – p) If X ~ B(10, 0.2) find the mean and variance of X E(X) = np Var(X) = np(1 – p)= 10 x 0.2 = 2 = 10 x 0.2 x 0.8= 1.6 If X ~ B(80, 0.4) find the mean and standard deviation of X E(X) = np Var(X) = np(1 – p)= 80 x 0.4 = 32 = 80 x 0.4 x 0.6= 19.2 = 4.38

11. X is a binomial distribution with mean 8 and variance 6.4. Find P(X ≤ 3). np = 8np(1 – p) = 6.4 8(1 – p) = 6.4 1 – p = 0.8p = 0.2 n = 40 P(X ≤ 3)= 0.0285

12. For the following random variables state whether they can be modelled by a binomial distribution. If they can, give the model, if they cannot then explain why. (a)A dice is thrown repeatedly until a 1 is seen. X = number of throws. (b)A dice is thrown 10 times. X = number of 1’s seen (c)A bag has 25 red and 25 blue balls in it. Five balls are taken out without replacement. X = number of red balls taken (d)X = number of boys in a family of five children (e)A pair of dice is thrown 25 times. X = number of times a double is thrown (f)A pair of dice is thrown 25 times. X = average score of the sum of the numbers showing.

13. (a)A dice is thrown repeatedly until a 1 is seen. X = number of throws.  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant NOT binomial No fixed number of trials

14. (b)A dice is thrown 10 times. X = number of 1’s seen  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant Binomial

15. (c) A bag has 25 red and 25 blue balls in it. Five balls are taken out without replacement. X = number of red balls taken  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant NOT binomial P(success) NOT constant Outcomes of trials NOT independent

16. (d)X = number of boys in a family of five children  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant Binomial

17. (e) A pair of dice is thrown 25 times. X = number of times a double is thrown  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant Binomial

18. (f)A pair of dice is thrown 25 times. X = average score of the sum of the numbers showing.  There is a fixed number of trials  Each trial has the same two possible outcomes  The outcomes of the trials are independent of one another  The probability of ‘success’ remains constant NOT binomial Different outcomes – not counting how many times something happens.

19. A vet thinks that the number of male puppies in litters of a Given size will follow a binomial distribution with p = 0.5. (a)In litters of six puppies, what would be the mean and variance of the number of males if the distribution is binomial? The vet records the number of males in 82 litters of six Puppies and the results are summarised in the table: (b)Calculate the mean and variance of the number of males in litters of six puppies. (c)Do you think the binomial distribution is a good model for the number of males in a litter of puppies? Males0123456 Frequency810161514127

20. A vet thinks that the number of male puppies in litters of a Given size will follow a binomial distribution with p = 0.5. (a)In litters of six puppies, what would be the mean and variance of the number of males if the distribution is binomial? E(X) = np Var(X) = np(1 – p)= 6 x 0.5 = 3 = 6 x 0.5 x 0.5= 1.5

21. The vet records the number of males in 82 litters of six Puppies and the results are summarised in the table: (b)Calculate the mean and variance of the number of males in litters of six puppies. Males0123456 Frequency810161514127

22. (c) Do you think the binomial distribution is a good model for the number of males in a litter of puppies? ModelActual Mean Variance 3 1.5 2.99 3.09 NO – Mean is a good match but variance is significantly bigger than model predicts.

23. (a) 40p(1 – p) = 9.640p – 40p 2 = 9.6 40p 2 – 40p + 9.6 = 0 If X ~ B(40, p) and Var(X) = 9.6 (a)Find the two possible values of p (b)For each of the values of p find P(X < μ – σ)

24. If X ~ B(40, p) and Var(X) = 9.6 (a)Find the two possible values of p (b)For each of the values of p find P(X < μ – σ) (b)Var(X) = 9.6σ = 3.098