 # Probability of Independent Events

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Probability of Independent Events

Probability of Independent Events
How is the probability of simple independent events determined? How is the probability of compound independent events determined?

The probability of an event is a number between
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 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 A will occur is: number of outcomes in A P (A) = 4 9 P (A) = total number of outcomes all possible outcomes outcomes in event A You can express a probability as a fraction, a decimal, or a percent. For example: , 0.5, or 50%. 1 2

1 6 Find the probability of rolling a 4. SOLUTION
Finding Probabilities of Events 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 1 6 = P (rolling a 4) = number of ways to roll the die

3 1 6 2 Find the probability of rolling an odd number. SOLUTION
Finding Probabilities of Events 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 3 6 1 2 = = P (rolling odd number) = number of ways to roll the die

Find the probability of rolling a number less than 7.
Finding Probabilities of Events 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 6 = = 1 P (rolling less than 7 ) = number of ways to roll the die

There are 52 cards in a deck. So what are my chances of picking an ace?

4 52 So I have a 4/52 or 1/13 chance of drawing an ace!
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 – 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 that are greater than 10 = 12, 14 = 2 total numbers – 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