Basic Rules of Probability
Section 3.3
The Axioms
If E is an event and is a subset of S, the sample
space, then the following are true:
1. 0 ≤ P(E) ≤1 The probability of an event E is
between 0 and 1 inclusive.
2. P() = 0 The probability of an empty set is
zero. Consequence: IF P(A ∩ B) = 0 then
A ∩ B =  which implies A and B are mutually
exclusive.
3. P(S)=1 The probability of the sample space is
1.
More Axioms
4. The rule for unions in general is
P(A U B) = P(A) + P(B) – P(A ∩ B)
5. If A and B are mutually exclusive then
P(A U B) = P(A) + P(B).
6.The rule for complimentary events is
P(E’) = 1 – P(E).
Examples
• Let E be the event of tossing two dice such that the sum
of the face is even. Let F be the event that the sum of
the faces is greater than 9.
• What is the probability of the sum of the faces being
even and greater than 9.
• What we are looking for is P(E ∩ F). The only rolls that
correspond to that event are (6,4), (4,6), (5,5), (6,6).
• Thus P(E ∩ F)=4/36=1/9.
• What is the probability of the sum being even or greater
than 9.
• Want to compute P(E U F).
• USE AN AXIOM P(E U F)=P(E)+P(F)-P(E ∩ F)=1/2+1/61/9=5/9
#61 p.155
• Of all flashlights in a large shipment, 15%
have a defective bulb, 10% have a
defective battery, and 5% have both
defects. If you purchase a flashlight from
the shipment what are the probabilities of
the following:
• A defective light bulb or a defective battery,
• A good bulb or a good battery,
• A good bulb and a good battery.
ANSWERS
•
•
•
•
•
•
•
•
•
•
L be the event of having a bad light bulb.
B be the vent of having a bad battery.
L’ is the event of having a good light bulb.
B’ is the event of having a good battery.
P(L)=0.15, P(B)=0.10, P(L ∩ B)=0.05
The probability of having a bad bulb or bad battery is P(L
U B)=P(L)+P(B)- P(L ∩ B)=
0.15+0.10-0.05=0.20=20%.
The probability of having a good bulb or good battery is
P(L’ U B’).
Using de’Morgan’s law L’ U B’=(L ∩ B)’
Thus P(L’ U B’) = P((L ∩ B)’)=1-P(L ∩ B)=1-0.05 = 0.95
= 95%
The last part where we have a good bulb and good
battery is P(L’∩B’)=P((L U B)’)=1-P(L U B)=1-.2=.8=80%