Download 3.2 Conditional Probability and the Multiplication Rule

3.2 Conditional Probability and
the Multiplication Rule
Conditional Probability
• Conditional Probability is the probability of
an event occurring, given that another
event has already occurred.
• The conditional probability of event B
occurring given that event A has occurred
is denoted by P (B A) and is read as
“probability of B, given A.
Finding Conditional Probabilities
Two cards are selected in sequence from a standard deck. Find the
probability that the second card is a queen, given that the first card is a king.
(Assume that the king is not replaced)
Because the first card is a king and is not replaced, the remaining deck has
51 cards and 4 are queens.
4
P (B A) =
≈ 0.078
51
Finding Conditional Probabilities
The table below shows the results of a study in which researchers examined a
child’s IQ and the presence of a specific gene in a child. Find the probability
that a child has a high IQ, given that the child has the gene.
Gene
present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
There are 72 children who have
the gene. So, the sample space
consists of these 72 children in
which 33 have a high IQ.
33
(
)
P BA =
≈ 0.458
72
Independent and Dependent
Events
• Two events are independent is the
occurrence of one of these events does
not affect the probability of the occurrence
of the other event.
• Two events A and B are independent if
P(B A) = P(B ) or if P( A B ) = P( A)
• Events that are not independent are
dependent
Classifying Events as Independent
or Dependent
• Selecting a king from a standard deck (A), not replacing
it, and then selecting a queen from the deck (B)
• The occurrence of A changes the probability of the
occurrence of B, so the events are dependent
• Tossing a coin and getting a head (A), then rolling a sixsided die and getting a 6 (B)
• The occurrence of A does not change the probability of
the occurrence of B, so the events are independent
• Driving over 85 miles per hour (A), and then getting into
a car accident (B)
• If you drive over 85 mile per hour, the chances of getting
into a car accident are greatly increased, so these
events are dependent
The Multiplication Rule
• The probability that two events A and B
will occur in sequence is
P( A and B ) = P( A) ⋅ P (B A)
• If event A and B are independent, then the
rule can be simplified to
P( A and B ) = P( A) ⋅ P(B )
The Multiplication Rule (in words)
• Find the probability that the first event
occurs
• Find the probability the second event
occurs given the first event has occurred
• Then multiply these two probabilities
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
A coin is tossed and a die is rolled. Find the probability of getting a head and
then rolling a 6.
The events are independent
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
A coin is tossed and a die is rolled. Find the probability of getting a head and
then rolling a 6.
The events are independent
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
A coin is tossed and a die is rolled. Find the probability of getting a head and
then rolling a 6.
The events are independent
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
Two cards are selected, without replacing the first card, from a standard deck.
Find the probability of selecting a king and then selecting a queen.
The events are dependent
P(K and Q ) = P(K ) ⋅ P (Q K )
4 4
= ⋅
52 51
16
=
≈ 0.006
2652
A coin is tossed and a die is rolled. Find the probability of getting a head and
then rolling a 6.
The events are independent
P(H and 6 ) = P(H ) ⋅ P(6 )
1 1
= ⋅
2 6
1
=
≈ 0.083
12
Using the Multiplication Rule
The probability that a salmon swims successfully through a dam is 0.85.
Find the probability that three salmon swim successfully through he dam.
One salmon's chance is independent of another
P(3 salmon are successful) = (0.85)(0.85)(0.85) ≈ 0.614
Find the probability that none of the three salmon are successful
Because the probabilty of success is 0.85, then the probability for failure is 0.15
P(none are successful) = (0.15)(0.15)(0.15) ≈ 0.003
Find the probability that at least one of the three salmon are successful
The phrase at least one means one or more. The compliment to the event is
none are successful.
P(at least 1 is successful) = 1 − P(none are successful)
≈ 1 − 0.003 ≈ 0.997