Department Store A department store is divided into two sections, electronics and furniture. Each section offers a discount rate; items in the same section sell at the same discount rate, but not necessarily at the same price. A.) One of the items in the electronics section has an original price $120 and a sale price of $102. One of the items in the furniture section has an original price of $60 and a sale price of $48. Determine the (percent) discount rate in the electronics section and the (percent) discount rate in the furniture section. Show your work. B.) Using the discount rate found in part A, find the total amount, including a 6% sales tax, that Andrew paid for an electric item and a furniture item if the original price of each of these items $200. Show your work. C.) Andrew plans to buy an office lamp from the furniture section and a laptop computer from the electronic section of this department store. The original price of the laptop computer is $1,350. The original price of the office lamp is $89. Andrew is thinking of two methods to find the total amount he will save (not including the tax). The two methods are stated below: Method I: Find the discount on the laptop and find the discount on the lamp. Find the sum of the two discounts Method II: Find the sum of the original prices of the two items. Find the average of the discount rates from part A Use the average rate to find the discount on the sum. Would Andrew’s results be the same using both methods? Show your work to explain your answer.
THEORETICAL AND EXPERIMENTAL PROBABILITY The probability of an event is a number between 0 and 1 that indicates the likelihood the event will occur. There are two types of probability: theoretical and experimental.
THEORETICAL AND EXPERIMENTAL PROBABILITY THE THEORETICAL PROBABILITY OF AN EVENT 4 P (A) = 9 total number of outcomes all possible outcomes You can express a probability as a fraction, a decimal, or a percent.For example: , 0.5, or 50%. 1 2 The theoretical probability of an event is often simply called the probability of the event. When all outcomes are equally likely, the theoretical probability that an event Awill occur is: number of outcomes in A P (A) = outcomes in event A
Finding Probabilities of Events 1 = 6 number of ways to roll the die You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling a 4. SOLUTION Only one outcome corresponds to rolling a 4. number of ways to roll a 4 P (rolling a 4) =
Finding Probabilities of Events 3 1 = = 2 6 number of ways to roll the die You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling an odd number. SOLUTION Three outcomes correspond to rolling an odd number: rolling a 1, 3, or a 5. number of ways to roll an odd number P (rolling odd number) =
Finding Probabilities of Events 6 = = 1 number of ways to roll the die 6 You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling a number less than 7. SOLUTION All six outcomes correspond to rolling a number less than 7. number of ways to roll less than 7 P (rolling less than 7 ) =
There are 52 cards in a deck. So what are my chances of picking an ace?
4 How many aces are in a deck? 52 How many cards are in a deck? So I have a 4/52 or 1/13 chance of drawing an ace!
When asked to determine the P(# or #) Mutually Exclusive Events • Mutually exclusive events cannot occur at the same time • Cannot draw ace of spaces and king of hearts • Cannot draw ace and king • But drawing a spade and drawing an ace are not mutually exclusive
Addition Rule for Mutually Exclusive Events • Add probabilities of individual events • Drawing ace of spades or king of hearts • Probability of ace of spades is 1/52 • Probability of king of hearts is 1/52 • Probability of either ace of spades or king of hearts is 2/52
Addition Rule for Not Mutually Exclusive Events • Add probabilities of individual events and subtract probabilities of outcomes common to both events
Drawing a spade or drawing an ace • Probability of drawing a spade: 13 outcomes, so 13/52 = 1/4 • Probability of drawing an ace: 4 outcomes, so 4/52 = 1/13 • Ace of spades is common to both events, probability is 1/52 • So probability of drawing a spade or an ace is 13/52 + 4/42 – 1/52 = 16/52 = 4/13
Independent and Dependent Events • Independent events: if one event occurs, does not affect the probability of other event • Drawing cards from two decks • Dependent events: if one event effects the outcome of the second event, changing the probability • Drawing two cards in succession from same deck without replacement
Multiplication Rule for Independent Events • To get probability of both events occurring, multiply probabilities of individual events • Ace from first deck and spade from second • Probability of ace is 4/52 = 1/13 • Probability of spade is 13/52 = 1/4 • Probability of both is 1/13 x 1/4 = 1/52
Conditional Probability • Probability of second event occurring given first event has occurred • Drawing a spade from a deck given you have previously drawn the ace of spade • After drawing ace of spades have 51 cards left • Remaining cards now include only 12 spades • Conditional probability is then 12/51
Probability Practice Problems • Suppose you have a bowl of disks numbered 1 – 15. • P(even) even numbers = 2, 4, 6, 8, 10, 12, 14 = 7 total numbers 1 – 15 = 15
Probability Practice Problems • Suppose you have a bowl of disks numbered 1 – 15. • P(even, more than 10) The “,” indicates “and” (the disk must be both even and more than 10) even #’s & #’s greater than 10 = 12, 14 = 2 total numbers 1 – 15 = 15
Probability Practice Problems • Suppose you have a bowl of disks numbered 1 – 15. • P(even or more than 10) The “or” indicates the disk must be even or more than 10. You must be careful not to include a number twice even #’s – 2, 4, 6, 8, 10, 12, 14 = 7/15 #’s greater than 10 = 11, 12, 13, 14, 15 = 5/15 Since 12 and 14 are common to both sets, you will subtract 2/15 7/15 + 5/15 – 2/15 = 10/15 = 2/3
Probability Practice Problems • Suppose you have a bowl of disks numbered 1 – 15. A disk is drawn, replaced, and a second disk is drawn. • P(even, even) Find the probability of each independent event and multiply even #’s on first draw – 2, 4, 6, 8, 10, 12, 14 = 7/15 even #’s on second draw – 2, 4, 6, 8, 10, 12, 14 = 7/15 7/15 x 7/15 = 49/225
Probability Practice Problems • Suppose you have a bowl of disks numbered 1 – 15. A disk is drawn, not replaced, and a second disk is drawn. • P(even, even) Find the probability of each independent event and multiply even #’s on first draw – 2, 4, 6, 8, 10, 12, 14 = 7/15 even #’s on second draw – one less even number than previous set/one less disk from bowl = 6/14 7/15 x 6/14 = 7/15 x 3/7 = 3/15 = 1/5
Independent Events Whatever happens in one event has absolutely nothing to do with what will happen next because: • The two events are unrelated OR • You repeat an event with an item whose numbers will not change (eg.: spinners or dice) OR • You repeat the same activity, but you REPLACE the item that was removed. The probability of two independent events, A and B, is equal to the probability of event A times the probability of event B. Slide 22
Independent Events P S O T R 6 1 5 2 3 4 Example: Suppose you spin each of these two spinners. What is the probability of spinning an even number and a vowel? P(even) = (3 evens out of 6 outcomes) (1 vowel out of 5 outcomes) P(vowel) = P(even, vowel) = Slide 23
Dependent Event • What happens the during the second event depends upon what happened before. • In other words, the result of the second event will change because of what happened first. The probability of two dependent events, A and B, is equal to the probability of event A times the probability of event B. However, the probability of event B now depends on event A. Slide 24
Dependent Event Example: There are 6 black pens and 8 blue pens in a jar. If you take a pen without looking and then take another pen without replacing the first, what is the probability that you will get 2 black pens? P(black first) = P(black second) = (There are 13 pens left and 5 are black) THEREFORE……………………………………………… P(black, black) = Slide 25
Tossing two dice and getting a 6 on both of them. • 2. You have a bag of marbles: 3 blue, 5 white, and 12 red. You choose one marble out of the bag, look at it then put it back. Then you choose another marble. • 3. You have a basket of socks. You need to find the probability of pulling out a black sock and its matching black sock without putting the first sock back. • 4. You pick the letter Q from a bag containing all the letters of the alphabet. You do not put the Q back in the bag before you pick another tile. TEST YOURSELF Are these dependent or independent events? Slide 26
Independent Events Find the probability • P(jack, factor of 12) x = Slide 27
Independent Events Find the probability • P(6, not 5) x = Slide 28
Dependent Events Find the probability • P(Q, Q) • All the letters of the alphabet are in the bag 1 time • Do not replace the letter x = 0 Slide 29
Probabilities of Dependent Events Determining probabilities of dependent events is usually more complicated than determining them for independent events. Since some of the tree diagrams could get very large, we will focus on a quicker method, multiplication.
Probabilities of Dependent Events Ex.1) Independent Events: Spinner #1 is partitioned into three equal sections, coloured black, white, and grey. Spinner #2 is partitioned into four equal sections, coloured red, blue, green, and yellow. If both spinners are spun, what is the probability of getting black and red? Since we expect to get black one-third of the time, and we expect to get red one-quarter of the time, then we expect to get black and red one-third of one-quarter of the time. . .
Probabilities of Dependent Events Imagine a tree diagram where the first column shows the three outcomes for Spinner #1, each of which is followed by the four outcomes for Spinner #2 in the second column. Three groups of four branches creates 12 possible outcomes.
Probabilities of Dependent Events Ex.2) Dependent Events: A bag contains 10 marbles; 5 red, 3 blue, and 2 silver. If you draw one marble at random and hold it in your left hand, and then draw a second marble at random and hold it in your right hand, what is the probability that you are holding two silver marbles? It’s easy to determine the probability of the first marble being silver. However, notice that if you start by getting a silver marble and then try for the second, the bag will be different. How? Now, there is only one silver marble in a bag containing a total of 9 marbles. . .