P(B) = Triangle Area P(A) = Oval Area P(A or B) = r + w + g P(A and B) = w So, P(A) + P(B) = r + w + g + w The Addition Rule for 2 events A and B P(A or B) = P(A) + P(B) - P( A and B)

What is the probability of choosing a red tile at random? = 2/9

Event: Five blue tiles are removed

What is the probability now, given the event? =1/2

A A B P(A|B) = P(A and B)/ P(B) Conditional Probability ABAB

An Example: Suppose 5% of 2m cars in a community are ‘bad’ (B), 20% are brand A and 2% are both bad and brand A (A and B). Question: What is the probability that a car is bad, given that it is brand A? Answer : |A| = ; |B| = |B and A| = P(B|A) = P(B and A)/P(A) =(40000/2m)/(400000/2m) = 0.1

Statistical independence Two events e and f are statistically independent if and only if P(e|f) = P(e) The probability of event B is dependent on the the event A. A and B are not independent events.

P (e | f) = P(e and f)/P( f) P (e and f ) = P (e) *P( f) The Multiplication rule: For two events e and f P (e and f) = P (e | f)* P ( f) If e and f are independent then

An example to show that two events can be mutually exclusive and dependent A bag contains 6 black and 4 red balls. A : Event the first ball is red Experiment : Two balls are chosen one after the other (with replacement) B: Event both are black

P(A) = 4/10 = 0.4 P(B) = 6/10 * 6/10 = 0.36 P(A and B) = 0 So A and B are mutually exclusive P(B|A) =P(A and B)/P(A) = 0 So, P(B|A)  P(B) A and B are not independent events But,

An example to show that two events can be independent yet not mutually exclusive A community has 1m people. 500,000 are adults (A) 600,000 are women (W) Obviously A and W are not mutually exclusive 300,000 adults are women (A and W)

P(A|W) = P(A and W)/P(W) = 300,000/600,000 = 0.5 Also, P(A) = 500,000/1 million = 0.5 Thus, P(A|W) = P(A) A and W are independent events

Arrangements and Selections Consider the following two problems: a. How many ways two letters can be selected from the set containing the three, A, B and C, (order of selection important) b. How many ways two letters can be selected from the set containing the three, A, B and C, (order of selection not important)

A B

A C

B A

B C

C A

C B

 In problem a, the answer is AB, AC, BA, BC, CA, CB.  The problem is relevant in choosing a password, for example.  The password AB123 is not the same as BA123 !  Each of the six possibilities is called an arrangement or a permutation.

 The 3 represents the number of letters to choose from and the number of terms in the multiplication represents the number of letters to choose. Altogether, we have 6 = 3 X 2 ways

 In problem b, the answer is AB, BC and CA, a total of 3 different selections or combinations.  This problem is relevant when selecting members of a committee, for example.  The committee consisting of member A being selected first and then B is not different from the one in which B was chosen first and then A.

AB AC BA CA BCCB AB BC AC Arrangements Selections  Now consider problem c:  How many ways three letters can be selected from the set containing the three, A, B, C. (order of selection important)?

A C B

A B C

B A C

B C A

C B A

C A B

ABC ACB BAC BCA CABCBA

 The 3 represents the number of letters to choose from and the number of terms in the multiplication represents the number of letters to choose. Altogether, we have 6 = 3 X 2 X 1 ways

 In general, if r letters are to be selected from n possible ones, where (r ≤ n), order of selection important, the total number of selections is n P r = nx(n-1)x(n-2)…x(n-r+1)  If r = n, we have the total number of selections = nx(n-1)x(n-2)….x2x1 = n! (read ‘factorial n’).  We cannot choose the same answer for problem b as that of a, since order is unimportant.

 AB is no different from BA!  So for every two selections in a, we need to take only one. Hence the answer is 3.

AB AC BA CA BCCB AB BC AC Arrangements Selections

 In general, if r letters are to be selected from n possible ones, where (r ≤ n), order of selection not important, we first start with the number  n P r = nx(n-1)x(n-2)…x(n-r+1)  Then we realise that the r letters selected can themselves be arranged in r! ways.  Since order is not important, we should just count only one selection for every r! of them.

 So the required number is  n C r = n P r /r! = {nx(n-1)x(n-2)…x(n- r+1)}/rx(r-1)x(r-2)x …x2x1  Thus, the number of 5-member committees formed out of 20 possible candidates is  20 C 5 = (20 x 19 x 18 x 17 x 16) / (5 x 4 x 3 x 2 x 1) =  Examples:

 1. A committee of three is to be selected from 3 boys and 4 girls. It must have exactly 1 boy and 2 girls in it. How many such committees can be formed?  Let the boys be called B1, B2 and B3, the girls G1, G2, G3 and G4 First, consider those selections in which B1 is selected. So two girls have to chosen to form the committee.

4  Two girls out of four can be chosen in 4 C 2 = 4x3/2x1 = 6 ways. So we shall have 6 committees in which B1 is a member.  But what about committees of which (i) B2 is a member and (ii) B3 is a member?  There are six of each type above.  So there are 6 x 3 = 18 committees altogether.

The formal way of answering the question is as follows: 31 boy can be chosen out of 3 in 3 C 1 = 3 possibilities. 4For each such selection of a boy, 2 girls out of 4 can be selected in 4 C 2 w= 6 different ways. 34So the total number of possible selections is 3 C 1 X 4 C 2 = 18

2. The UK National Lottery (i) What is the probability of winning the jackpot from a single ticket? (ii) What is the probability of winning exactly £10 from a single ticket? 49Answer (i) : If 6 balls are chosen out of 49 then there are 49 C 6 or ways of doing so.

The machine can select any one of combinations. So P(one ticket wins jackpot) = 1/ Answer (ii): Consider the six (yet unknown) numbers to be chosen by the machine and call them ‘winners’. So there are 6 winners and 43 losers. A ticket wins £10 if it selects exactly 3 winners out of 6 and 3 losers out of 43.