 # © Boardworks Ltd 20051 of 32 © Boardworks Ltd 2005 1 of 32 AS-Level Maths: Statistics 1 for Edexcel S1.3 Probability This icon indicates the slide contains.

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© Boardworks Ltd 20051 of 32 © Boardworks Ltd 2005 1 of 32 AS-Level Maths: Statistics 1 for Edexcel S1.3 Probability This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation.

© Boardworks Ltd 20052 of 32 Contents © Boardworks Ltd 2005 2 of 32 Probability Probability basics and notation Estimating probability Addition properties Independent events Conditional probability

© Boardworks Ltd 20053 of 32 Uncertainty is a feature of everyday life. Probability is an area of maths that addresses how likely things are to happen. A good understanding of probability is important in many areas of work. It is used by scientists, governments, businesses, insurance companies, betting companies and many others, to help them anticipate future events. How likely am I to live to 100? Probability Am I likely to win the lottery? Are interest rates likely to go up? Which team is most likely to win the FA cup?

© Boardworks Ltd 20054 of 32 A statistics experiment will have a number of different outcomes. The set of all possible outcomes is called the sample space of the experiment. Introduction to probability An event is a collection of some of the outcomes from an experiment. For example, getting an even number on the dice or scoring more than 40 on the quiz. In a general knowledge quiz with 70 questions, the sample space for the number of questions a person answers correctly is {0, 1, 2, …, 70}. if a normal dice is thrown the sample space would be {1, 2, 3, 4, 5, 6}. For example:

© Boardworks Ltd 20055 of 32 Let A be an event arising from a statistical experiment. The probability that A occurs is denoted P(A) (where 0 ≤ P(A) ≤ 1). Notation The probability that A does not occur is denoted P(A′). If A is impossible, then P(A) = 0. If A is certain to happen, then P(A) = 1. P(A′) = 1 – P(A)

© Boardworks Ltd 20056 of 32 When two experiments are combined, the set of possible outcomes can be shown in a sample space diagram. Introduction to probability 6789101112 567891011 45678910 3456789 2345678 1234567 123456 First throw Second throw P(total = 6) = There are 36 equally likely outcomes. Example: A dice is thrown twice and the scores obtained are added together. Find the probability that the total score is 6. 5 of the outcomes result in a total of 6. This notation means “probability that the total = 6”.

© Boardworks Ltd 20057 of 32 Contents © Boardworks Ltd 2005 7 of 32 Probability basics and notation Estimating probability Addition properties Independent events Conditional probability Estimating probability

© Boardworks Ltd 20058 of 32 Some probabilities are less simple. It is not always possible to calculate how likely each outcome is. Estimating probability You can increase the accuracy of the relative frequency as an estimate of probability, by increasing the number of times you repeat the experiment. This is referred to as the relative frequency. However, the probability of an event happening can be estimated experimentally, by repeating an experiment over and over again. The probability is estimated using:

© Boardworks Ltd 20059 of 32 Estimating probability

© Boardworks Ltd 200510 of 32 Venn diagrams Venn diagrams can be used to represent probabilities. The outcomes that satisfy event A can be represented by a circle. A The outcomes that satisfy event B can be represented by another circle. B The circles can be overlapped to represent outcomes that satisfy both events.

© Boardworks Ltd 200511 of 32 Contents © Boardworks Ltd 2005 11 of 32 Probability basics and notation Estimating probability Addition properties Independent events Conditional probability Addition properties

© Boardworks Ltd 200512 of 32 In Venn diagrams representing mutually exclusive events, the circles do not overlap. Two events A and B are called mutually exclusive if they cannot occur at the same time. A B If A and B are mutually exclusive, then: Addition properties This symbol means ‘union’ or ‘OR’ However the events “the card is a club” and “the card is a queen” are not mutually exclusive. For example, if a card is picked at random from a standard pack of 52 cards, the events “the card is a club” and “the card is a diamond” are mutually exclusive.

© Boardworks Ltd 200513 of 32 This addition rule for finding P(A  B) is not true when A and B are not mutually exclusive. P(A  B) = P(A) + P(B) – P(A  B) Addition properties The more general rule for finding P(A  B) is: This symbol means ‘intersect’ or ‘AND’ Venn diagrams can help you to visualize probability calculations.

© Boardworks Ltd 200514 of 32 Example: A card is picked at random from a pack of cards. Find the probability that it is either a club or a queen or both. Addition properties Card is a club = event C Card is a queen = event Q So, This area represents the 12 clubs that are not queens. This represents the queen of clubs. This represents the other 3 queens.

© Boardworks Ltd 200515 of 32 Example 2: If P(A′  B′) = 0.1, P(A) = 0.45 and P(B) = 0.75, find P(A  B). We can deduce that: P(A  B) = 1 – 0.1 = 0.9 A B 0.1 Using the formula, P(A  B) = P(A) + P(B) – P(A  B), we get: 0.9 = 0.45 + 0.75 – P(A  B) 0.9 = 1.2 – P(A  B) Addition properties P(A′  B′) is the unshaded area in the Venn Diagram. So, P(A  B) = 0.3

© Boardworks Ltd 200516 of 32 Examination-style question: There are two events, C and D. P(C) = 2P(D) = 3P(C  D). Given that P(C  D) = 0.52, find: a) P(C  D) b) P(C  D′). a)Let P(C  D) = x P(C  D) = P(C) + P(D) – P(C  D) So, 0.52 = 3 x + 1.5 x – x Therefore x = P(C  D) = 0.52 ÷ 3.5 = 0.149 (3 s.f.) Addition properties C D x

© Boardworks Ltd 200517 of 32 Question (continued): b)P(C  D′) corresponds to the unshaded area in this Venn diagram. We see that: P(C  D′) = P(C′  D′) + P(C) = (1 – 0.52) + 3 × 0.149 = 0.48 + 0.447 = 0.927 (3 s.f.) …as P(C) = 3P(C  D) Addition properties C D

© Boardworks Ltd 200518 of 32 Contents © Boardworks Ltd 2005 18 of 32 Probability basics and notation Estimating probability Addition properties Independent events Conditional probability Independent events

© Boardworks Ltd 200519 of 32 Two events are said to be independent if the occurrence of one has no effect on the probability of the second occurring. If A and B are independent, then: Independent events P(A  B) = P(A) × P(B) For example, if a coin and a die are both thrown, then the events “the coin shows a head” and “the die shows an odd number” are independent events.

© Boardworks Ltd 200520 of 32 Example: A and B are independent events. P(A) = 0.7 and P(B) = 0.4. a) Find P(A  B). b) Find P(A′  B). A B Independent events a) As A and B are independent, P(A  B) = P(A) × P(B) = 0.7 × 0.4 = 0.28 b)P(A′  B) is the shaded region in the Venn diagram. So, P(A′  B) = P(B) – P(A  B) = 0.4 – 0.28 = 0.12

© Boardworks Ltd 200521 of 32 Tree diagrams are sometimes a useful way of finding probabilities that involve a succession of events. Independent events Example: A bag contains 6 green counters and 4 blue counters. A counter is chosen at random from the bag and then replaced. This is repeated two more times. Find the probability that the 3 counters chosen are: a)all green b)not all the same colour.

© Boardworks Ltd 200522 of 32 G G G G G G G B B B B B B B 0.6 0.4 0.6 0.4 0.6 0.4 a)P(GGG) = 0.6 × 0.6 × 0.6 Independent events = 0.216

© Boardworks Ltd 200523 of 32 b)To find the probability that there will be at least one of each colour, we can find 1 – P(GGG) – P(BBB) P(BBB) = 0.4 × 0.4 × 0.4 = 0.064. Independent events G G G G G G G B B B B B B B 0.6 0.4 0.6 0.4 0.6 0.4 Therefore, the answer is 1 – 0.216 – 0.064 = 0.72

© Boardworks Ltd 200524 of 32 Contents © Boardworks Ltd 2005 24 of 32 Probability basics and notation Estimating probability Addition properties Independent events Conditional probability

© Boardworks Ltd 200525 of 32 However, the probability of event B happening might depend on whether A has happened or not. For example, if blue and green counters are pulled from a bag twice and not replaced, then the probability of pulling out a green counter on the second try will depend on what colour was pulled out on the first try. Conditional probability The probability that event A will happen, given that event B has happened, is written P(A | B) This is a conditional probability.

© Boardworks Ltd 200526 of 32 Conditional probability B A A′A′ B B′B′ B′B′ P(A) P(A′) P(B | A) P(B′ | A) P(B | A′ ) P(B′ | A′ ) P(A  B) = P(A) × P(B | A) P(A  B′) = P(A) × P(B′ | A) P(A′  B′) = P(A′) × P(B′ | A′) P(A′  B) = P(A′) × P(B | A′) To find the probability of events A and B both happening we use: P(A  B) = P(A) × P(B | A) This formula can be re-arranged to give:

© Boardworks Ltd 200527 of 32 D M D D M M Example: A bag contains 8 dark chocolates and 4 milk chocolates. One chocolate is taken out and eaten. A second chocolate is then taken. Find the probability that: a) two milk chocolates are taken. b) the two chocolates are of different types. Conditional probability a) P(M  M) = b) P(D  M) = P(M  D) =

© Boardworks Ltd 200528 of 32 Examination-style question: A man has 2 shirts (one white and one blue) and 2 ties (red and silver). If he wears the white shirt, he chooses the red tie with probability 0.4. If he wears the blue shirt, he chooses the red tie with probability 0.75. The probability that he wears the white shirt is 0.7. Conditional probability a)Find the probability that he wears the red tie. b)Given that he is wearing a red tie, find the probability that he picked the blue shirt.

© Boardworks Ltd 200529 of 32 W B R R S S a)P(red tie) = P(W  R) + P(B  R) = (0.7 × 0.4) + (0.3 × 0.75) = 0.28 + 0.225 = 0.505 Shirt Tie 0.7 0.3 0.4 0.6 0.75 0.25 Conditional probability

© Boardworks Ltd 200530 of 32 So, b)Recall the formula for conditional probability: Conditional probability = 0.446 (3 s.f.)

© Boardworks Ltd 200531 of 32 Example: A disease affects 1 in 500 people. A diagnostic test for the disease records a positive result 99% of the time when the disease is present (this is called the sensitivity of the test). The test records a negative result 95% of the time when the disease in not present. The test results are always either positive or negative. Find the probability that a person has the disease, given that the test result is positive. The example below demonstrates an important application of probability in the field of medicine. Conditional probability

© Boardworks Ltd 200532 of 32 D D′D′ +ve –ve DiseaseTest 0.002 0.998 0.99 0.01 0.05 0.95 P(D  +ve) = 0.002 × 0.99 = 0.00198 P(D′  +ve) = 0.998 × 0.05 = 0.0499 Conditional probability Therefore, P(+ve) = 0.05188 So,

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