Simple Probability
The probability of event A occurring is the total number of ways event A can happen
divided by the total number of outcomes. A probability can be a fraction, a decimal or a
percent. However the probability of a A occurring, written P(A) must be as below.
0  P( A)  1
Where P(A) = 1 means A will certainly happen and P(A) = 0 means that A is impossible.
Example 1: A card is drawn randomly from a deck of playing cards. Event A is picking a Red King. Find
the P(A).
Example 2: A fair, six sided die is thrown. Event E is getting a even number less than five. Find P(E).
“OR” Type Probabilities
The probability that events A or B happens is
P(A or B) = P(A) + P(B) – P(A and B)
Often, P(A and B) = 0 and we only have to look at P(A) + P(B).
Example 2: Draw one card randomly from a deck of playing cards. What is the probability of drawing a
King (event A) OR drawing an Red card (event B)?
Solution:
P(A or B) = P(A) + P(B) – P(A and B)
Independent Events
We say that two events A and B are independent if the occurrence of one of the events
does not effect the probability of the other event occurring. Let’s exam two events.
A = you wear a blue shirt to school
B = you are late for school
Obviously, these two events are “independent”, that is the color of your shirt will not
effect whether you are late to school and vice versa. Now let’s look at two other events.
A = you can’t find your keys in the morning
B = you are late for school
In this case A and B are not independent or as we say, A and B are “dependent”.
“AND” Type Probabilities
The probability that event A and event B both occur is
P(A and B) = P(A) P(B/A)
where the P(B/A) means that the probability of B occurring given that A has already
occurred. The “/” means “given”. If A and B are independent events, the equation is
much simpler
P(A and B) = P(A) P(B)
so we usually hope that our events are independent.
Example 1: We take the example of an LSUA student who is real mess. He has a terrible
wardrobe and is forced to wear a blue shirt (event A) to school 1/5 of the time. He’s late
to school (event B) 1/3 of the time and given that he can’t find his car keys (event C),
he’s late to school (event B) half the time. Not only that, but he can’t find his car keys
(event C) ¼ of the time.
Find P(A and B)
Solution:
List events and probabilities.
A= wears a blue shirt to school, P(A) = 1/5
B = late for school. P(B) = 1/3
C = can’t find his car keys. P(C) = 1/4
B/C = late for school given he can’t find his car keys. P(B/C) = 1/2