Transcript Probability-of-Independent-Events

Probability of
Independent Events
Probability of Independent
Events Essential Questions
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 and
experimental.
THEORETICAL AND EXPERIMENTAL PROBABILITY
THE THEORETICAL PROBABILITY OF AN EVENT
The theoretical
probability
of anlikely,
eventthe
is
When
all outcomes
are equally
often simplyprobability
called the probability
of theA event.
theoretical
that an event
will occur is:
P (A) =
number of outcomes in A
total number of outcomes
all possible
outcomes
P (A) = 4
9
outcomes
in event A
You can express a probability as a fraction, a decimal, or a percent.
For example: 1 , 0.5, or 50%.
2
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
P (rolling a 4) =
number of ways to roll the die
1
=
6
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.
P (rolling odd number) =
number of ways to roll an odd number
3
=
=
number of ways to roll the die
6
1
2
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.
P (rolling less than 7 ) =
number of ways to roll less than 7
6
=
=1
number of ways to roll the die
6
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 & #’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