Download S-300 4-2D AandB Depnd

Section 4 – 2D: !
Conditional Probability Involving Dependent Events
Dependent Events
P (A AND B )
A bag contains several marbles. You randomly draw one marble from the bag, record itʼs color, and
then you DO NOT REPLACE the first marble drawn. You then randomly draw a second marble
from the bag and record itʼs color. When you DO NOT REPLACE the first marble the contents of
the bag changes after the first draw. One of the marbles has been taken out in Event A and NOT
Replaced. If you are given the outcome of Event A then the new sample space for Event B can be
listed and the probability for Event B can be found.
If the outcome of the second event (Event B) depends on the outcome of the first event (Event A)
then we say that the two events are Dependent. The probability of a second marble being selected
is NOT the P(B) used in the independent case. The probability of a specific second marble being
selected depends on what marble was selected in Event A. If the outcome of the second event
(Event B) depends on the outcome of the first event (Event A) then probability of Event B is stated as
“the probability of B given that A has happened”. This is written as P(B | A)
The original Sample Space is used to find the probability of Event A or P(A). After the outcome of
Event A is given then a new sample space is listed and the the probability of B given that A has
happened is found using the new sample space.
Section 4 – 2D!
Page 1 of 11!
© 2016 Eitel
Example 1
You randomly draw 2 marbles, one at a time WITHOUT replacement.
The two events are dependent.
Find P( R and B ) WITHOUT replacement
Find the probability that you get a Red marble and then a Blue marble WITHOUT replacement.
At the start of the problem a bag contains 18 marbles { 8R , 2 B }
First Event: You draw a Red marble WITHOUT replacement.
The bag now has only 9 marbles left in it and there is one less Red marble.
The bagʼs contents!
at the start of the!
first selection!
The bagʼs contents
at the start of the
second selection
Solution:
P( R and B ) WITHOUT replacement = P(R) • P(B | R)
The first marble is selected from 10 marbles { 8R , 2B }
P(R) =
8 possible Red marbles
8
=
10 total marbles
10
The second draw is different. The bag now has only 9 marbles left in it and there is one less Red
marble. The second marble is selected from 9 marbles { 7R , 2B }.
P(B | R) =
2 possible Blue marbles
2
=
9 total marbles
9
P( R and G ) WITHOUT replacement = P(R) • P(G | R) =
Section 4 – 2D!
8 2
8
2/ 1
8
• =
•
=
5
10 9
9
45
1/ 0/
Page 2 of 11!
© 2016 Eitel
Example 2
You randomly draw 2 marbles, one at a time WITHOUT replacement.
The two events are dependent.
Find P( G and B ) WITHOUT replacement
Find the probability that you get a Green marble and then a Blue marble WITHOUT replacement.
At the start of the problem the a contains 15 marbles { 2R , 3 G , 10 B }
Event A: You draw a Blue marble WITHOUT replacement.
The second bag now has only 14 marbles left in it and there is one less Blue marble.
The bagʼs contents!
at the start of the!
first selection!
The bagʼs contents
at the start of the
second selection
Solution:
P( G and B ) WITHOUT replacement = P(G) • P(B | B)
The first marble is selected from 10 marbles
P(G) =
{ 2R , 3 G , 10 B }
3 possible green marbles
3
=
15 total marbles
15
The second draw is different. The bag now has only 14 marbles left in it and there is one less Green
marble. The second marble is selected from 14 marbles { 2R , 2 G , 10 B }
P(B | B) =
10 possible Blue marbles
10
=
14 total marbles
14
P( G and B ) WITHOUT replacement = P(G) • P(B | B) =
Section 4 – 2D!
Page 3 of 11!
3 10
3
9/ 3
9
• =
•
=
= .13
5
15 14
14
70
1/ 5/
© 2016 Eitel
Example 3
You randomly draw 2 marbles, one at a time WITHOUT replacement.
The two events are dependent.
Find P( Y and G ) WITHOUT replacement
Find the probability that you get a Yellow marble and then a NOT Green marble WITHOUT
replacement.
At the start of the problem the a contains 10 marbles { 3R , 6 G , 1 Y }
Event A: You draw a Yellow marble WITHOUT replacement.
The second bag now has only 9 marbles left in it and there is one less Yellow marble.
The bagʼs contents!
at the start of the!
first selection!
The bagʼs contents
at the start of the
second selection
Solution:
P( Y and G ) without replacement = P(Y) • P( G | Y)
The first marble is selected from 10 marbles { 3R , 6 G , 1 Y }
P(Y) =
1 possible yellow marble
1
=
10 total marbles
10
The second draw is different. The bag now has only 9 marbles left in it and there is one less yellow
marble. The second marble is selected from 9 marbles { 3R , 6G, 0Y }.
P( G | Y) =
3 possible NOT Green marbles
3
=
9 total marbles
9
P( Y and G ) without replacement = P(Y) • P( G | Y) =
Section 4 – 2D!
1 3
1 3/ 1
1
• =
• 3 =
= .03
10 9
1/ 0 9/
30
Page 4 of 11!
© 2016 Eitel
Example 4
You randomly draw 2 marbles, one at a time WITHOUT replacement.
The two events are dependent.
Find P( Y and Y ) WITHOUT replacement
Find the probability that you get a NOT Yellow marble and then a NOT Yellow marble WITHOUT
replacement.
At the start of the problem the a contains 12 marbles { 4R , 6 G , 2 Y }
Event A: You draw a NOT Yellow marble WITHOUT replacement.
The second bag now has only 9 marbles left in it and there is one less NOT Yellow marble.
Not Yellow means Red or Green. One of the 10 Red or Green marbles was taken in the first draw
so 9 Red or Green marbles remain.
The bagʼs contents!
at the start of the!
first selection!
!
Solution:
The bagʼs contents
at the start of the
second selection
R + G = 10
R+G=9
P( Y and Y ) WITHOUT replacement = P( Y) • P( Y | Y)
The first marble is selected from 12 marbles { 4R , 6 G , 2 Y }
P( Y) =
10 possible not yellow marbles
12
=
12 total marbles
10
The second draw is different. The bag now has only 9 marbles left in it and there is one less NOT
yellow marble. The second marble is selected from 11 marbles { 9 Red or Green, 0Y }.
P( Y | Y) =
9 possible NOT Yellow marbles
9
=
11 total marbles
11
P( Y and Y ) WITHOUT replacement = P( Y) • P( Y | Y ) =
Section 4 – 2D!
Page 5 of 11!
10 9
1/ 0/ 5 9
45
• =
•
=
= .68
6
12 11 1/ 2/
11
66
© 2016 Eitel
Example 5
You randomly draw 3 marbles, one at a time WITHOUT replacement.
The three events are dependent.
Find P( 3 B ) WITHOUT replacement
Find the probability that you get a Blue marble and then a Blue marble and then a Blue Marble
WITHOUT replacement.
At the start of the problem the a contains 12 marbles { 4R , 6 G , 2 Y }
Event A: You draw a Blue marble WITHOUT replacement.
The bagʼs contents
at the start of the
first selection
The bagʼs contents The bagʼs contents
at the start of the!
second selection!
at the start of the
third selection
Solution:
P( B and B and B ) WITHOUT replacement = P(B) • P(B | B) • P(B | 2B)
The first marble is selected from 15 marbles { 2R , 3 G , 10 B }
P(B) =
10 possible Blue marbles
10
=
15 total marbles
15
The second marble is selected from 14 marbles { 2R , 3 G , 9 B }
P(B | B) =
9 possible Blue marbles
9
=
14 total marbles
14
The third marble is selected from 13 marbles { 2R , 3 G , 8 B }
P(B | 2B) =
8 possible Blue marbles
8
=
13 total marbles
13
10 9 8
1/ 0/ 2
9/ 3 8/ 4
24
P(3B ) WITHOUT replacement =
• • =
•
=
= .26
1 •
7
/
15 14 13 1/ 5/ 3 1/ 4/
13
91
Section 4 – 2D!
Page 6 of 11!
© 2016 Eitel
Example 7
A Bag contains 6 marbles { 2G , 4Y } You randomly draw 2 marbles, one at a time without
replacement. Find all possible probabilities.
Section 4 – 2D!
Page 7 of 11!
© 2016 Eitel
Optional Extra Examples
P (A AND B ) for Dependent Events
Example 1
A Bag contains 9 marbles { 2Y , 6B, 1G } You randomly draw 2 marbles, one at a time without
replacement.
Find the probability that you get a Green marble then a Blue marble. P(G and B )
Solution:
You randomly draw 2 marbles, one at a time without replacement. The 2 events are dependent.
P( G and B ) = P(G) • P(B | G)
The first marble is selected from!
9 marbles { 2Y , 6B, 1G } !
P(G) =
The second marble is selected from
8 marbles { 2Y , 6B, 0G }
1 possible Green marble
1
=
!
9 total marbles
9
P( B and G ) = P(B) •P(B | G) =
P(B | G) =
6 possible BLUE marbles
6
=
8 total marbles
8
1 6
1 6/ 3
3
1
• =
• 4 =
=
9 8
9 8/
36 12
Example 2
A Bag contains 7 marbles { 2G , 5Y } You randomly draw 2 marbles, one at a time without
replacement. Find the probability that you get a Green marble then a Green marble. P(G and G)
Solution:
You randomly draw 2 marbles, one at a time without replacement. The 2 events are dependent.
P( G and G ) = P(G) • P(G | G)
The first marble is selected from!
7 marbles { 2G , 5Y } !
P(G) =
The second marble is selected from
6 marbles { 1G , 5Y }
2 possible Green marbles
2
=
!
7 total marbles
7
P(G | G) =
1 possible Green marbles
1
=
6 total marbles
6
2/ 1 1
1
• 3=
21
P( G and G ) = P(G) • P(G | G) = 7 6/
Section 4 – 2D!
Page 8 of 11!
© 2016 Eitel
Example 3
A Bag contains 7 marbles { 1R , 6B } You randomly draw 2 marbles, one at a time without
replacement. Find the probability that you get a RED marble then a Blue marble. P(R and B )
Solution:
You randomly draw 2 marbles, one at a time without replacement. The 2 events are dependent.
P( R and B ) = P(R) •P(B | R)
The first marble is selected from!
The second marble is selected from
7 marbles { 1R , 6B } !
P(R) =
1 possible Red marble
1
=
7 total marbles
7
P( R and B ) = P(R) •P(B | R) =
6 marbles { 0R , 6B }
!
P(B | R) =
6 possible Blue marbles
6
= =1
6 total marbles
6
1 6
1
• =
7 6
7
Example 4
A Bag contains 4 marbles { 1R , 3B } You randomly draw 2 marbles, one at a time without
replacement. Find the probability that you get a RED marble then a Red marble. P(R and R )
Solution:
You randomly draw 2 marbles, one at a time without replacement. The 2 events are dependent.
P( R and R ) = P(2R) = P(R) • P(R | R)
The first marble is selected from!
The second marble is selected from
4 marbles { 1R , 3B } !
P(R) =
1 possible Red marble
1
=
4 total marbles
4
P( R and R ) = P(R) • P(R | R) =
Section 4 – 2D!
3 marbles { 0R , 3B }
!
P(R | R) =
0 possible Red marbles
0
=
3 total marbles
3
1 0
• = 0
4 3
Page 9 of 11!
© 2016 Eitel
Example 5
A Bag contains 7 marbles { 4R , 3B } You randomly draw 3 marbles, one at a time without
replacement. Find the probability that you get 3 RED marbles. P(R and R and R )
Solution:
You randomly draw 3 marbles, one at a time without replacement. The 3 events are dependent.
P( R and then a R and then a R ) = P( 3R ) = P(R) •P(R | R) • P(R | 2R)
The first marble is selected from!
7 marbles { 4R , 3B } !
P(R) =
4 possible Red marbles
4
=
7 total marbles
7
The second marble is selected from
6 marbles { 3R , 3B }
P(R | R) =
!
3 possible Red marbles
3
=
6 total marbles
6
The third marble is selected from!
5 marbles { 2R , 3B } !
P(R | 2R) =
2 possible Red marbles
2
=
5 total marbles
5
P( 3R ) = P(R) • P(R | R) • P(R | 2R) =
Section 4 – 2D!
!
4 3 2
4
• • =
7 6 5
35
Page 10 of 11!
© 2016 Eitel
Example 6
A Bag contains 9 marbles { 4R , 2B , 3Y} You randomly draw 2 marbles, one at a time without
replacement. Find the probability that you get a Yellow marble and then a Not Red marble.
Solution:
You randomly draw 2 marbles, one at a time without replacement. The 2 events are dependent.
P( Y and R ) = P(Y) • P( R | Y)
The first marble is selected from!
9 marbles { 4R , 2B , 3Y} !
P(Y) =
The second marble is selected from
8 marbles { 4R , 2B , 2Y}
3 possible Yellow marbles
1
=
!
9 total marbles
3
P( Y and R ) = P(Y) • P( R | Y) =
Section 4 – 2D!
P( R | Y) =
4 possible NOT Red marbles
1
=
8 total marbles
2
1 1
1
• =
3 2
6
Page 11 of 11!
© 2016 Eitel