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) = Pr(S) = Pr(S) = =

• 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)

BNot B (B complem

ent)

A

Not A (A complem

ent)Pr(B) + Pr(B’)

= 1

Pr(A) + Pr(A’) = 1

• 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 0.5A’ 0.2

1

This box should always add to ONE

HOW CAN WE USE KARNAUGH MAPS?

0.4 0.60.30.3

0.5

HOW CAN WE USE KARNAUGH MAPS?

• From this, we can answer questions:

• What is Pr(A ∩ B’)? • Pr(A ∩B’) = 0.3• Refer back to the template to find the different probabilities

B B’A 0.2 0.3 0.5A’ 0.2 0.3 0.5

0.4 0.6 1

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 cheeseChicken

Not chicken 10.6

7

0.240.2

30.33

0.76

0.10.53

0.14

Find the probability that the randomly chosen burger:• a) Has cheese AND chicken

Pr(cheese ∩ chicken) = 0.14• 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

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

QUESTIONS TO DO

•Complete the Karnaugh maps worksheet