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AS-Level Maths: Statistics 1 for OCR MEI 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. 11 of of 32 32 © Boardworks Ltd 2005 Probability basics and notation Contents Probability basics and notation Estimating probability Addition properties Independent events Conditional probability 22 of of 32 32 © Boardworks Ltd 2005 Probability How likely am I to live to 100? Which team is most likely to win the FA cup? Are interest rates likely to go up? Am I likely to win the lottery? 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. 3 of 32 © Boardworks Ltd 2005 Introduction to probability A statistics experiment will have a number of different outcomes. The set of all possible outcomes is called the sample space of the experiment. For example: if a normal dice is thrown the sample space would be {1, 2, 3, 4, 5, 6}. 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}. 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. 4 of 32 © Boardworks Ltd 2005 Notation Let A be an event arising from a statistical experiment. The probability that A occurs is denoted P(A) (where 0 ≤ P(A) ≤ 1). If A is certain to happen, then P(A) = 1. If A is impossible, then P(A) = 0. The probability that A does not occur is denoted P(A′). P(A′) = 1 – P(A) 5 of 32 © Boardworks Ltd 2005 Introduction to probability When two experiments are combined, the set of possible outcomes can be shown in a sample space diagram. There are 36 equally likely outcomes. 5 of the outcomes result in a total of 6. P(total = 6) = 5 36 Second throw Example: A dice is thrown twice and the scores obtained are added together. Find the probability that the total score is 6. 6 7 8 9 10 11 12 5 6 7 8 9 10 11 4 5 6 7 8 9 10 3 4 5 6 7 8 9 2 3 4 5 6 7 8 1 2 3 4 5 6 7 1 2 5 6 3 4 First throw This notation means “probability that the total = 6”. 6 of 32 © Boardworks Ltd 2005 Estimating probability Contents Probability basics and notation Estimating probability Addition properties Independent events Conditional probability 77 of of 32 32 © Boardworks Ltd 2005 Estimating probability Some probabilities are less simple. It is not always possible to calculate how likely each outcome is. However, the probability of an event happening can be estimated experimentally, by repeating an experiment over and over again. The probability is estimated using: number of times event occurs number of times experiment is repeated This is referred to as the relative frequency. You can increase the accuracy of the relative frequency as an estimate of probability, by increasing the number of times you repeat the experiment. 8 of 32 © Boardworks Ltd 2005 Estimating probability 9 of 32 © Boardworks Ltd 2005 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. 10 of 32 © Boardworks Ltd 2005 Addition properties Contents Probability basics and notation Estimating probability Addition properties Independent events Conditional probability 11 11 of of 32 32 © Boardworks Ltd 2005 Addition properties Two events A and B are called mutually exclusive if they cannot occur at the same time. 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. However the events “the card is a club” and “the card is a queen” are not mutually exclusive. If A and B are mutually exclusive, then: P(A B ) = P(A) + P(B ) A B In Venn diagrams representing mutually exclusive events, the circles do not overlap. This symbol means ‘union’ or ‘OR’ 12 of 32 © Boardworks Ltd 2005 Addition properties This addition rule for finding P(A B) is not true when A and B are not mutually exclusive. The more general rule for finding P(A B) is: P(A B) = P(A) + P(B) – P(A B) This symbol means ‘intersect’ or ‘AND’ Venn diagrams can help you to visualize probability calculations. 13 of 32 © Boardworks Ltd 2005 Addition properties 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. Card is a club = event C Card is a queen = event Q 1 P(C ) = 4 4 1 P(Q ) = = 52 13 P(C Q ) = This represents the other 3 queens. This area represents the 12 clubs that are not queens. 1 52 This represents the queen of clubs. 1 1 1 4 + – = So, P(C Q ) = 4 13 52 13 14 of 32 © Boardworks Ltd 2005 Addition properties Example 2: If P(A′ B′) = 0.1, P(A) = 0.45 and P(B) = 0.75, find P(A B). P(A′ B′) is the unshaded area in the Venn Diagram. A B We can deduce that: 0.1 P(A B) = 1 – 0.1 = 0.9 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) So, P(A B) = 0.3 15 of 32 © Boardworks Ltd 2005 Addition properties 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′). C a) Let P(C D) = x D x P(C D) = P(C) + P(D) – P(C D) So, 0.52 = 3x + 1.5x – x Therefore x = P(C D) = 0.52 ÷ 3.5 = 0.149 (3 s.f.) 16 of 32 © Boardworks Ltd 2005 Addition properties Question (continued): C D 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 …as P(C) = 3P(C D) = 0.48 + 0.447 = 0.927 (3 s.f.) 17 of 32 © Boardworks Ltd 2005 Independent events Contents Probability basics and notation Estimating probability Addition properties Independent events Conditional probability 18 18 of of 32 32 © Boardworks Ltd 2005 Independent events Two events are said to be independent if the occurrence of one has no effect on the probability of the second occurring. 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. If A and B are independent, then: P(A B) = P(A) × P(B) 19 of 32 © Boardworks Ltd 2005 Independent events Example: A and B are independent events. P(A) = 0.7 and P(B) = 0.4. a) Find P(A B). A B b) Find P(A′ B). 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 20 of 32 © Boardworks Ltd 2005 Independent events Tree diagrams are sometimes a useful way of finding probabilities that involve a succession of 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. 21 of 32 © Boardworks Ltd 2005 Independent events 0.6 0.4 0.6 0.4 0.4 0.6 B G 0.4 0.6 B G 0.4 B 0.6 G 0.4 B B G B 0.4 G G G 0.6 0.6 B a) P(GGG) = 0.6 × 0.6 × 0.6 = 0.216 22 of 32 © Boardworks Ltd 2005 Independent events 0.6 0.4 0.6 0.4 0.4 0.6 B G 0.4 0.6 B G 0.4 B 0.6 G 0.4 B B G B 0.4 G G G 0.6 0.6 B 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. Therefore, the answer is 1 – 0.216 – 0.064 = 0.72 23 of 32 © Boardworks Ltd 2005 Conditional probability Contents Probability basics and notation Estimating probability Addition properties Independent events Conditional probability 24 24 of of 32 32 © Boardworks Ltd 2005 Conditional probability 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. The probability that event A will happen, given that event B has happened, is written P(A | B) This is a conditional probability. 25 of 32 © Boardworks Ltd 2005 Conditional probability To find the probability of events A and B both happening we use: P(A B) = P(A) × P(B | A) P(B | A) B P(A B) = P(A) × P(B | A) P(B′ | A) B′ P(A B′) = P(A) × P(B′ | A) P(B | A′ ) B P(A′ B) = P(A′) × P(B | A′) P(B′ | A′ ) B′ P(A′ B′) = P(A′) × P(B′ | A′) A P(A) P(A′) A′ This formula can be re-arranged to give: P(A B ) P(B | A) P(A) 26 of 32 © Boardworks Ltd 2005 Conditional probability 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. 2 3 1 3 27 of 32 7 11 D 4 11 M 8 11 D 3 11 M D M 1 3 1 a) P(M M) = 3 11 11 2 4 8 b) P(D M) = 3 11 33 1 8 8 P(M D) = 3 11 33 8 8 16 33 33 33 © Boardworks Ltd 2005 Conditional probability 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. 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. 28 of 32 © Boardworks Ltd 2005 Conditional probability Tie Shirt 0.4 R 0.6 S 0.75 R 0.25 S W 0.7 0.3 B a) P(red tie) = P(W R) + P(B R) = (0.7 × 0.4) + (0.3 × 0.75) = 0.28 + 0.225 = 0.505 29 of 32 © Boardworks Ltd 2005 Conditional probability b) Recall the formula for conditional probability: P(A B ) P(B | A) P(A) P(B R ) 0.225 So, P(B | R ) P(R ) 0.505 = 0.446 (3 s.f.) 30 of 32 © Boardworks Ltd 2005 Conditional probability The example below demonstrates an important application of probability in the field of medicine. 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. 31 of 32 © Boardworks Ltd 2005 Conditional probability Disease Test 0.99 +ve 0.01 –ve 0.05 +ve 0.95 –ve D 0.002 0.998 D′ P(D +ve) = 0.002 × 0.99 = 0.00198 P(D′ +ve) = 0.998 × 0.05 = 0.0499 Therefore, P(+ve) = 0.05188 P(D +ve) 0.00198 0.0382 (3 s.f.) So, P(D |+ve) P(+ve) 0.05188 32 of 32 © Boardworks Ltd 2005