1. Conditional probabilities and independence of events

2. Calcul des probabilities One of the classic problems in probability theory, which offers us an introduction to the notion of conditional probabilities, was posed by the French mathematician, Joseph Louis Francois Bertrand, in his textbook Calcul des probabilities (1889) Bertrand's Box Problem

3. Two coins in each of 3 boxes There are three boxes: a box with two gold coins, a box with two silver coins, and a box with one coin of each kind. A box is chosen at random and a coin is removed. What is the probability that the other coin in that box also is gold?

4. A counter-intuitive answer To many people it may seem that the probability of the chosen box's other coin being gold is 1/2, but this is not the case – it is actually 2/3. Initially, before any box is selected, it is true that the probability is ½ that the second coin will be gold, but then “extra” information about the first coin taken from that box alters this probability, because some of the initial outcomes no longer are realistic possibilities

5. A sample-space diagram G1G2G3S1S2S3G1G2G3S1S2S3 G 2 = 1/6 G 1 = 1/6 S 1 = 1/6 G 3 = 1/6 S 3 = 1/6 S 2 = 1/6 First coin Second coin probability

6. The reduced sample-space G1G2G3S1S2S3G1G2G3S1S2S3 G 2 = 1/3 G 1 = 1/3 S 1 = 1/3 G 3 = 1/6 S 3 = 1/6 S 2 = 1/6 First coin Second coin probability

7. Venn Diagram E F E∩FE∩F Event E: the 2nd coin is gold (3 sample-points) Event F: the 1st coin is gold (3 sample points Event E∩F: both coins are gold (2 sample points) Conditional probability: P( E│F ) = P( E∩F ) / P( F ) = (2/6) / (3/6) = 2/3 S

8. Independent Events independent Two events E and F are independent if the extra knowledge that event F has occurred does not alter the probability of event E occurring In symbols: P( E│F ) = P( E )

9. A coin-tossing example Suppose a coin is tossed two times. The probability of getting HEADS on the 2 nd toss is unaffected by getting HEADS on the 1 st toss. First toss Second Toss Probability H = 1/4 H T = 1/4 H = 1/4 T T = 1/4

10. An example with dice Suppose a player tosses a pair of dice. Let's consider two events: Event E: the faces agree Event F: the sum is even Is event E independent of event F? (i.e., does P(E) = P(E│F) ? )

11. Sample space 1 2 3 4 5 6 654321654321 Event E: the faces agree (probability = 6/36) Event F: the sum is even (probability = 18/36) P( E │ F ) = P( E∩F ) / P( F ) = (6/36) / (18/36) = 6/18 ≠ P( E )

12. Observation using algebra From the definition of conditional probability: P( E│F ) = P( E∩F ) / P( F ) and from the definition of independent events: P( E│F ) = P( E ) symmetrical condition there follows this symmetrical condition for events E and F which are independent: P( E∩F ) = P( E )P( F )

13. Multinomial coefficients The number of ways of depositing n distinct objects into k distinct bins, with m objects in the 1 st bin, m objects in the 2 nd bin, and so on, where n = m 1 + m 2 +... + m k, multinomial coefficient Is given by the multinomial coefficient n n! m 1 m 2... m k m 1 ! m 2 !... m k ! () =

14. Example 1 In a class of 27 recruits, a navy instructor assigns letter-grades A, B, C, D, F according to the military handbook's prescribed ratio: 1:2:3:2:1, which means that 3 will get A's, 6 will get B's, 9 will get C's, 6 will get D's, and 3 will get F's. The number of different grade-assignments possible can be expressed by this multinomial coefficient multinomial coefficient: 27 3 6 9 6 3 The value of this multinomial coefficient is given by the ratio: 27! / 3!6!9!6!3! ()

15. Example 2 A school principal must assign 90 students to three classrooms, each of which has 30 built-in desks. The number of possible assignments can be expressed as a multinomial coefficient: 90 30 30 30 ( ) This multinomial coefficient can be evaluated as a ratio: 90! / 30!30!30!. Alternatively, this problem can be solved by regarding the principal's job as consisting of a sequence of three tasks, in which 30 students are chosen for the first classroom, then 30 are chosen for the second classroom, and finally the remaining 30 students are assigned to the third classroom. By the Fundamental Counting Principle, the total number of assignments can be expressed as a product of three binomial coefficients: (90 choose 30) x (60 choose 30) x (30 choose 30)

16. Probability example A business woman with 7 keys wants to keep 2 of them in her briefcase, and 2 of them in her purse; the rest go into her car's glove-compartment. If two of these keys are for operating her automobile, but she distributes the keys randomly to those three locations, then what is her liklihood of putting both car-keys in the same location?

17. solution A business woman with 7 keys wants to keep 2 of them in her briefcase, and 2 of them in her purse; the rest go into her car's glove-compartment. If two of these keys are for operating her automobile, but she distributes the keys randomly to those three locations, then what is her liklihood of putting both car-keys in the same location? 5 5 5 0 2 3 2 0 3 2 2 1 7 2 2 3 ) ( ( ) + ( ) (

18. Choosing a committee In a survey of class attendance an instructor finds that, of 14 students present, 5 are Freshmen, 5 are Sophomores, and 4 are Juniors. If he wants to select a 5-member committee consisting of 2 Freshmen, 2 Sophomores, and 1 Junior, then how many ways are possible for this to be done? Notice that this is NOT a problem about multinomial coefficients. Fundamental Counting Principle Here we can fall back on the Fundamental Counting Principle, by organizing a sequence of separate tasks that will produce the desired committee, counting the numbers of ways each separate task can be done, then multiplying those numbers together. First: select 2 Freshman from the 5 who are present ( b(5;2) = 10 ways ) Next: select 2 Sophomores from the 5 who are present (again, 10 ways) Last: select 1 Junior from the 4 who are present ( b(4;1) = 4 ways) So the total number of possible committees is the product: (10)(10)(4) = 400