PROBABILITY – KARNAUGH MAPS
WHAT IS A KARNAUGH MAP?
• A Karnaugh map is a table which shows the probabilities involved in Venn diagrams or
two-way tables
• A typical Venn diagram shows how many outcomes fall in each circle, like this:
• We can then find the probabilities of these events occurring, for example
Pr(S) =
𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑤ℎ𝑜 𝑐ℎ𝑜𝑠𝑒 𝑠𝑝𝑎𝑔ℎ𝑒𝑡𝑡𝑖
𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑠𝑢𝑟𝑣𝑒𝑦𝑒𝑑
11+ 9
Pr(S) = 11+9+15+5
Pr(S) =
20
40
=
1
2
• A Karnaugh map can be used to show the probabilities of these events occurring in table form
IN A VENN DIAGRAM:
• A is everything in the ‘A’ circle
• B is everything in the ‘B’ circle
• A ∩ B’ means ‘A but not B’
• A ∩ B means ‘A and B’ or ‘overlap/intersection of A and B’
• A’ ∩ B means ‘B but not A’
• A’ ∩ B’ means ‘not A or B’ or ‘everything outside the A and B circles’
IN A KARNAUGH MAP:
•
•
We can set these probabilities out in table form
The rows and columns can be added up, for example, if we look at the first column:
Pr(A ∩ B) + Pr(A’ ∩ B) = Pr(B)
B
Not B (B
complement)
A
Not A (A
complement)
Pr(B) + Pr(B’) = 1
Pr(A) + Pr(A’) = 1
HOW CAN WE USE KARNAUGH MAPS?
• They are useful if we have some values out of the table, because we can use
them to fill in the gaps
B
B’
A
0.2
0.3
0.5
A’
0.2
0.3
0.5
0.4
0.6
1
This box should always
add to ONE
HOW CAN WE USE KARNAUGH MAPS?
• From this, we can answer questions:
B
B’
A
0.2
0.3
0.5
A’
0.2
0.3
0.5
0.4
0.6
1
• What is Pr(A ∩ B’)?
• Pr(A ∩B’) = 0.3
• Refer back to the template to find the different probabilities
WORDED PROBLEM USING KARNAUGH MAPS
•
You go to a restaurant where they sell different types of burgers. The probability of
choosing a burger at random and getting one with cheese is 0.67, getting a burger
with chicken is 0.24, and not getting cheese or chicken is 0.23
•
•
•
•
•
Find the probability that the randomly chosen burger:
a) Has cheese and chicken
b) Has cheese or chicken
c) Has no cheese
d) Has chicken but no cheese
FILLING IN THE KARNAUGH MAP TABLE
• You go to a restaurant where they sell different types of burgers.
The
probability of choosing a burger at random and getting one with cheese is
0.67, getting a burger with chicken is 0.24, and not getting cheese or chicken
is 0.23
Cheese
Not cheese
Chicken
0.14
0.1
0.24
Not chicken
0.53
0.23
0.76
0.67
0.33
1
Find the probability that the randomly chosen burger:
• a) Has cheese AND chicken
Pr(cheese ∩ chicken) = 0.14
Remember the addition law of probability:
Pr(A U B) = Pr(A) + Pr(B) – Pr(A ∩ B)
You will need this formula to help figure out unknown
values in some Karnaugh maps
• b) Has cheese OR chicken
Pr (cheese U chicken) = Pr(cheese) + Pr(chicken) – Pr(cheese ∩ chicken) = 0.67 + 0.24 – 0.14 = 0.77
• c) Has no cheese
Pr(cheese’) = 0.33
• d) Has chicken but no cheese
Pr(chicken ∩ cheese’) = 0.1
QUESTIONS TO DO
• Complete the Karnaugh maps worksheet