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?

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 andexperimental.

THEORETICAL AND EXPERIMENTAL PROBABILITY

THE THEORETICAL PROBABILITY OF AN EVENT

When all outcomes are equally likely, thetheoretical probability that an event A will occur is:

P (A) = total number of outcomes

The theoretical probability of an event is often simply called the probability of the event.

all possible outcomes

number of outcomes in A

outcomes in event A

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

P (A) = 49

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.

P (rolling a 4) = number of ways to roll a 4

number of ways to roll the die16

=

Finding Probabilities of Events

Three outcomes correspond to rolling an odd number: rolling a 1, 3, or a 5.

P (rolling odd number) = number of ways to roll an odd number

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

36

12

= =

Finding Probabilities of Events

All six outcomes correspond to rolling a number less than 7.

P (rolling less than 7 ) = number of ways to roll less than 7

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

number of ways to roll the die66

= = 1

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

How many aces are in a deck?

How many cards are in a deck?

4

52

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 that are 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