Ronald Hui Tak Sun Secondary School HKDSE Mathematics Ronald Hui Tak Sun Secondary School

Homework SHW7-B1: Sam L SHW7-P1: Sam L RE8: Sam L EHHW1: Tashi, Kelvin, Sam L, Pako EHHW2: Tashi, Kelvin, Sam L, Pako Ronald HUI

Homework SHW9-B1: Sam L SHW9-C1: Sam L SHW9-R1: Tashi, Sam L SHW9-P1: Tashi, Sam L RE9: Sam L UT2: Sam L Ronald HUI

Homework SHW10-A1: Sam L SHW10-B1: Sam L SHW10-C1: Sam L SHW10-R1: Sam L, Matthew(RD), Ken(RD), Ronald(RD), Charles(RD), Macro S(RD) SHW10-P1: Sam L RE10: Sam L SQ10: Sam L Ronald HUI

Q:\School Publishing\Secondary (Maths)\Cathy\NSS MIA2nd\5-Min Lecture\Eng\5B11_5Min_05e.ppt Conditional Probability and Multiplication Law of Probability for Dependent Events

Concept of Conditional Probability An urn contains 2 blue balls and 1 red ball. Two balls are drawn one by one at random without replacement: Event A: The first ball drawn is blue. Event B: The second ball drawn is blue. If event A has occurred, there are 1 red ball and 1 blue ball left. The probability of event B given that event A has occurred is

Concept of Conditional Probability An urn contains 2 blue balls and 1 red ball. Two balls are drawn one by one at random without replacement: Event A: The first ball drawn is blue. Event B: The second ball drawn is blue. If event A did not occur (i.e. the first ball drawn is red), there are 0 red balls and 2 blue balls left. The probability of event B given that event A did not occur is 1.

In some situations, the occurrence of an event would affect the probability of the other event. These kinds of probabilities are called conditional probabilities. In symbolic form, for any two events A and B, the probability of event B given that event A occurred is denoted by P(B | A).

The table shows the number of students in class 5A and class 5B. Q:\School Publishing\Secondary (Maths)\Cathy\NSS MIA2nd\5-Min Lecture\Eng\5B11_5Min_05e.ppt 5A 5B Boy 15 9 Girl 12 10 The table shows the number of students in class 5A and class 5B. A student is chosen at random from the classes. Find the probabilities that the student is a girl given that the student is from 5B. the student is from 5B given that she is a girl. 10 of them satisfy the event ‘the student is from 5B ’. 10 of them satisfy the event ‘the student is a girl’. (a) P(girl | 5B) (b) P(5B | girl) There are 9 + 10 possible outcomes for the condition ‘the student is from 5B’. There are 12 + 10 possible outcomes for the condition ‘the student is a girl’. Note: From the above example, we observe that P(girl | 5B)  P(5B | girl). In general, P(B | A)  P(A | B).

Follow-up question A fair coin is tossed 3 times. (a) List all the possible outcomes. (b) Find the probability of getting exactly one tail given that there is at least one tail. (a) The possible outcomes are: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (b) There are 7 possible outcomes for the condition ‘at least one tail’. 3 of them satisfy the event ‘exactly one tail’. ∴ P(exactly one tail | at least one tail)

Multiplication Law of Probability for Dependent Events Ken takes MTR to school everyday. Event A: There is a train failure in the MTR line which Ken takes. Event B: Ken is late for school. Does the occurrence of event A affect the probability that event B occurs? Yes / No The occurrence of event A affects the probability that event B occurs. Events A and B are dependent events.

An urn contains 3 blue balls and 2 red balls. Two balls are drawn one by one at random without replacement: The first ball drawn is blue. Event C: The second ball drawn is also blue. Event D: Are events C and D dependent? Yes, P(the 2nd ball drawn is blue) is affected by the outcome of the first draw.

In throwing a dice twice: The number ‘1’ is got in the first throw. Event A: The number ‘6’ is got in the second throw. Event B: Are events A and B dependent? No, P(the second number is ‘6’) is not affected by the outcome of the first throw.

Let us explore the relationships among P(A), P(B | A) and P(A  B) in the following example.

What is the relationship among them? Consider four cards numbered 1, 2, 3 and 4. In drawing two cards at random one by one without replacement: Sample space = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)} Experiment P(A) P(B|A) P(A  B) P(A)  P(B | A) (i) Event A: The 1st number is ‘1’. Event B: The 2nd number is ‘3’. (ii) Event A: The 1st number is even. Event B: The 2nd number is prime. Are the above events, A and B, dependent in each of the above case? Yes / No What is the relationship among them? P(A  B) = P(A) P(B | A) 

if A and B are two dependent events, then P(A  B) = P(A)  P(B | A). In general, if A and B are two dependent events, then P(A  B) = P(A)  P(B | A). In fact, this relationship is the multiplication law of probability for dependent events. Note that if A and B are two independent events, then P(B | A) = P(B). i.e. P(A  B) = P(A)  P(B)

What is the probability that the pens drawn are both red? Consider two red pens and two blue pens in a box. Two pens are drawn one by one without replacement. What is the probability that the pens drawn are both red?

First draw: P(the first one is red) Second draw: P(the second one is red  the first one is red) P(both are red) = P(the first one is red)  P(the second one is red  the first one is red) =  1 2 3 P(A  B) = P(A)  P(B | A)

Follow-up question There are 4 blue balls and 3 red balls in an urn. Jay randomly takes out two balls one by one from the urn without replacement. What is the probability that the balls taken are both red? P(both are red) = P(the first one is red)  P(the second one is red  the first one is red)

conditional probability P(B | A) Using Formula to find Conditional Probabilities We learnt that if A and B are two dependent events, then P(A  B) = P(A)  P(B | A). For P(A) > 0, the conditional probability P(B | A) can be found by , where P(A) > 0

There are 40 students in a laboratory There are 40 students in a laboratory. 32 students are wearing a laboratory coat and 10 of them are wearing a pair of safety goggles also. A student is selected randomly from the laboratory. Find the probability that the student is wearing a pair of safety goggles given that he/she is wearing a laboratory coat. P(wearing a pair of safety goggles | wearing a laboratory coat) = P(wearing a pair of safety goggles  wearing a laboratory coat) P(wearing a laboratory coat)

Follow-up question In a class, 40% of the students are girls and 30% of the students are girls with glasses. A student is selected randomly from the class. Find the probability that the student is wearing glasses given that she is a girl. = P(wearing glasses  girl) P(girl) P(wearing glasses | girl)

Using Permutation and Combination to Solve Probability Problems Q:\School Publishing\Secondary (Maths)\Cathy\NSS MIA2nd\5-Min Lecture\Eng\5B11_5Min_06e.ppt Using Permutation and Combination to Solve Probability Problems

In this section, we’ll see how to use permutation and combination to solve probability problems.

Take the election of student representatives as an example. Q:\School Publishing\Secondary (Maths)\Cathy\NSS MIA2nd\5-Min Lecture\Eng\5B11_5Min_06e.ppt Take the election of student representatives as an example. Suppose there are 5 candidates and each of them has equal chance of being elected as a student representative. A B C D E If 2 student representatives are to be elected, the order in which the 2 candidates are selected is not important. Then, the number of possible ways is C 5 2 = 10. Therefore, the probability that candidates A and B are elected is C 5 2 1 10 1 = Selecting A and B is one of the possible ways. C 5 2

Take the election of student representatives as an example. Q:\School Publishing\Secondary (Maths)\Cathy\NSS MIA2nd\5-Min Lecture\Eng\5B11_5Min_06e.ppt Take the election of student representatives as an example. Suppose there are 5 candidates and each of them has equal chance of being elected as a student representative. A B C D E However, if a President and a Treasurer are to be elected from the 5 candidates, the order in which the 2 candidates are selected is important. P 5 2 Then, the number of possible ways is = 20. Therefore, the probability that candidates A and B are elected as the President and Treasurer respectively is P 5 2 1 20 1 =

Follow-up question In a game, there are 10 balls numbered from 0 to 9 in a lottery box. The host of the game draws three balls from the lottery box one by one without replacement. A player has to guess the three numbers on the balls drawn. Find the probabilities that the player guesses all the three numbers drawn by the host correctly in order, (b) none of the three numbers guessed by the player is drawn by the host.

Number of ways of drawing 3 balls from 10 balls in which the order is important = P 10 3 P 10 3 1 720 1 = ∴ The required probability = = C 10 3 (b) Number of ways of drawing 3 balls from 10 balls Number of ways that none of the three numbers guessed by the player is drawn by the host = C 7 3 ∴ The required probability C 10 3 = 7 24 7 =