# PROBABILITY OF INDEPENDENT AND DEPENDENT EVENTS SECTION 12.5.

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PROBABILITY OF INDEPENDENT AND DEPENDENT EVENTS SECTION 12.5

WHEN ASKED TO DETERMINE THE 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

PROBABILITY OF INDEPENDENT EVENTS If A and B are independent events, then the probability that both A and B occur is P(A and B) = P(A) ● P(B)

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

S T R O P 1 2 3 6 5 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) = Independent Events Slide 9

PROBABILITY PRACTICE PROBLEMS

EX: A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips. One paperclip is taken from the drawer and then replaced. Another paperclip is taken from the drawer. What is the probability that the first paperclip is red and the second paperclip is blue? P(red then blue) = P(red) P(blue) = 3/12 5/12 = 15/144 = 5/48.

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. If A and B are dependent events, the probability of both events occurring is the product of the probability of the first event and the probability of the second event once the first event has occurred. Slide 12

If A and B are dependent events, and A occurs first, P(A and B) = P(A) P(B,once A has occurred)... and is written as... P(A and B) = P(A) P(B|A) The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A) [pronounced as The probability of event B given A]. The notation used above does not mean that B is divided by A. It means the probability of event B given that event A has already occurred. To find the probability of the two dependent events, we use a modified version of Multiplication Rule

EXAMPLE: A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips. One paperclip is taken from the drawer and is NOT replaced. Another paperclip is taken from the drawer. What is the probability that the first paperclip is red and the second paperclip is blue? Because the first paper clip is NOT replaced, the sample space of the second event is changed. The sample space of the first event is 12 paperclips, but the sample space of the second event is now 11 paperclips. The events are dependent. P(red then blue) = P(red) P(blue) = 3/12 5/11 = 15/132 = 5/44.

EXAMPLE: There are 3 red candies left in a bag of multicolored candies with a total of 20 candies left in it. The probability that you will get a red one when you reach in is: 3/20. But what are your chances of getting a red one if you reach in again? There are now 19 candies in the bag, and only two are red. The probability is 2/19. Taking the first candy affected the outcome of the next attempt. The two events are dependent.

4. Four cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing a ten, a nine, an eight and a seven in order? 5. Three cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing 3 aces?

Determine whether the events are independent or dependent. 1. selecting a marble and then choosing a second marble without replacing the first marble 2. rolling a number cube and spinning a spinner Two fair dice, one red and one blue, are rolled. Find each probability. 3. P(1 and 1) 4. P(4 and 1) 5. P(an even number and a 3) 6. P(a number greater than 4 and a 2) 7. Standardized Test Practice David and Adrian have a coupon for a pizza with one topping. The choices of toppings are pepperoni, hamburger, sausage, onions, bell peppers, olives, and anchovies. If they choose at random, what is the probability that they both choose hamburger as a topping?

Determine whether the events are independent or dependent. 1. selecting a marble and then choosing a second marble without replacing the first marble 2. rolling a number cube and spinning a spinner Two fair dice, one red and one blue, are rolled. Find each probability. 3. P(1 and 1) 4. P(4 and 1) 5. P(an even number and a 3) 6. P(a number greater than 4 and a 2) 7. Standardized Test Practice David and Adrian have a coupon for a pizza with one topping. The choices of toppings are pepperoni, hamburger, sausage, onions, bell peppers, olives, and anchovies. If they choose at random, what is the probability that they both choose hamburger as a topping?