RULES OF PROBABILITY COMPLEMENTARY EVENTS: Consider any event A. Let p(a) be the probability that A happens and let p(a ) read as the probability of A prime or A c (A Complement), be the probability that A does not happen. Therefore, the p A) + p (A ) =1. In conclusion, we can say that the probability of A not happening is 1 p (A), or p (A c ) = 1 p (A). OR The COMPLEMENT OF A is elements that are NOT in Complement of an event: p (A c ) = 1 p (A) EXAMPLE 1. Using the events below, what is the complement of EVENT A? EVENT A = {1, 2, 3, 4, 5, 7, 9}; EVENT B = {3, 4, 6, 8, 9}; P (A c ) = {6, 8} P (A c ) =The Complement of an EVENT A are the elements that are NOT IN EVENT A but only elements in EVENT B. TRY THESE 1. 2. Two fair coins are tossed. What is the probability at least one coin lands heads up? P (h) = 1 p (t c ) 3. A fair die is thrown. What is the probability that the score is not a factor of 6? P (not a factor of 6) = 1 p (factors of 6) 4. Two fair dice are thrown. What is the probability that the two scores do not add to 7? 5. A spinner is made from a piece of card in the shape of a regular pentagon with a toothpick pushed through the center. When the spinner is spun and it lands on an edge, each of the numbers from 1 to 5 is equally likely. If the spinner is spun twice, what is the probability the two scores are different?
ADDITION RULE: (GENERAL CASE); If A and B are any events, the probability of obtaining either of them is equal to the probability of A plus the probability of B minus the probability of both occurring at the same time. Symbolically, the addition rule is as follows. If A and B are any events, then: p (A or B) = p (A) + p (B) p (A and B) We can also manipulate the equation: p (A and B) = -p (A or B) +p (A) + p (B) EXAMPLE 1 OF ADDITION RULE: The probability that any member of the Hamilton Bay Ensemble plays a guitar is 1/4, and the probability that a member plays a clarinet is 5/8. If the probability that a member plays both instruments is 5/24, what is the probability that the member plays the guitar or that the member plays the clarinet? SOLUTION: Let Event A = Playing the guitar; EVENT B = Playing the clarinet; Probability the guitar or the clarinet = p (A or B) USING THE ADDITION RULE P (plays guitar or clarinet) = p (plays guitar) + p (plays clarinet) p (plays guitar and clarinet) Using the ADDITION RULE: P (A or B) = p (A) + p (B) p (A and B) = 1/4 + 5/8 5/24 = 6/24 + 15/24 5/24 = 16/24 =2/3 EXAMPLE 2 OF ADDITION RULE: Amanda spins a spinner numbered 1 to 10. What is the probability that the spinner lands on an odd number that is divisible by 3? Sample space: = 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 Let Event A = p (odd number) = {1, 3, 5, 7, 9}; = 5/10 Let EVENT B =p (# divisible by 3) = {3, 6, 9} = 3/10; P (A and B) = {3, 9} = 2/10 Did you notice that elements 3 and 9 are common to the list of odd numbers and also divisible by 3? Therefore, we should apply the ADDITION RULE here: P (odd number or #divisible by 3) = p (odd #) + p (# divisible by 3) - p (odd # and # divisible by 3) P (A or B) = p (A) + p (B) p (A and B) P (a or B) = 5/10 + 3/10 2/10; = 8/10 2/10 = 6/10 = 3/5
Addition Rule of Probability - Step-by-Step Lesson An ice cream cup has 3 scoops of chocolate, 4 scoops of vanillas and 3 scoops of butter scotch ice cream. What is the probability that if you randomly sampled the ice cream that you will taste chocolate or vanilla ice cream? Explanation: Step 1) First we have to see what we have to find out. Step 2) It can be expressed as: P(chocolate) = 3/10; P(vanillas) = 4/10; P(chocolate) + P(vanillas) = 3/10 + 4/10 = 7/10 NOW IT S YOUR TURN: 1. Find the probability that if you roll two six-sided dice, you get doubles or a sum of 8. 2. Find the probability that if you roll two six-sided dice, you get doubles or a sum of 4. 3. Find the probability that if you roll two six-sided dice, you get doubles or a sum of 7. 4. Find the probability that if you roll two six-sided dice, you get a 5 on the first die or get a 5 on the second die. 5. Find the probability that if you roll a pair of dice, you do not get doubles or you get a sum of 8.
6. Find the probability that if you flip two fair coins, you get heads on the first coin or you get heads on the second coin. 7. In a local school, 80% of the students carry a backpack (B) or a wallet (W). Forty percent carry a backpack, and 50% carry a wallet. If a student is selected at random, find the probability that the student carries both a backpack and a wallet. 8. Suppose 80% of people can swim. Suppose 70% of people can whistle. Suppose 55% of people can do both. What percentage of people can swim or whistle? ****************************************************************************** MULTIPLICATION RULE: The probability that event A and event B both happen is given by: P (A and B) = P (A)* P (B/A); Alternatively, P (A and B) = P (B) * P (A/B) CONDITIONAL PROBABILITY Although the addition rule given above applies to many different situations, there are still other problems that cannot be solved by that formula. It is for this reason that we introduce conditional probability. The definition of conditional probability is that for any two events A and B, where P (B) > 0, the probability of A given the condition B is P (A/B) = P (A and B)/P (B). A ( A and B) p ( ) p provided p (B) 0 B p (B) EXAMPLE 1 of CONDITIONAL PROBABILITY RULE: In Ashville the probability that a married man drives is 0.90. If the probability that a married man and his wife both drive is 0.85, what is the probability that his wife drives given that he drives? SOLUTION: We apply the conditional probability formula above: Let A = wife drives; Let B = husband drives. Therefore, p (A/B) = p(a and B)/p(B)
We are told that; p (husband drives) = 0.90; and P (husband and wife drive) = 0.85 (husband and wife drive) 0.85 85 17 Thus, p (wife drives/husband drives) = p 94.4%. p (husband drives) 0.90 90 18 EXAMPLE 2 OF CONDITIONAL PROBABILITY RULE: The students in a second grade classroom were surveyed to see if they have any pets. Of the students surveyed, 68% of them said they have a dog and 17% said they have a dog and a cat. What is the probability that a student has a cat given the student already has a dog? Let cat = Event A; Let dog be Event B P (A/B) = p (A and B)/ p(b). Event B = 0.68 (student has a dog); Event A and B = 0.17 (student has a cat and dog); (0.17) 17 1 P (student has a cat and a dog)/ p (student has a dog); p or.25 or 25% (0.68) 68 4 EXAMPLE 3 OF CONDITIONAL PROBABILITY RULE: In a school of 1200 students, 250 are seniors, 150 students take math, and 40 students are seniors and are also taking math. What is the probability that a randomly chosen student who is a senior, is taking math? Answer: These questions can be confusing. It sounds, at first read, that they are asking for the probability of choosing a student who is a senior and who is taking math. Not quite right! It helps to re-word the question into: Find the probability that the student is taking math, given that the student is a senior. B = the student is taking math n(a) = the student is a senior = 250. n(a and B) = the student is a senior and is taking math = 40.
NOW IT S YOUR TURN: 9. Joe often speeds while driving to school in order to arrive on time. The probability that he will speed to school is 0.75. If the probability that he speeds and gets stopped by a police officer is 0.25, find the probability that he is stopped given he is speeding. 10. Janet likes to study. The probability that she studies and passes her math test is 0.80. If the probability that she studies is 0.83, what is the probability that she passes the math test, given she has studied? 11. A survey conducted in school shows that 56% of the boys like dogs and 32% of the boys like both dogs and cats. What approximate percentage of boys like cats given that they also like dogs?