1. Probability Problems Solved with
Combinations

2. Basics
Three cards are drawn from a well-shuffled standard deck of 52 cards, one after the other and without replacement.
Find the probability of drawing
all clubs b) no clubs
exactly one club (hint: the club can occur on the 1st, 2nd or 3rd drawing)
Evaluate: b)
Compare your answers to exercises 1 and 2. What do you notice?
Write and evaluate an expression using combinations to find the probability of getting exactly 2 clubs.
They are the same!

3. Solving Probability Problems w/ Combinations
Recall the definition of probability:
If E is an event from a sample space S of equally likely outcomes, the probability of event E is:
0  P(E)  1.
Counting techniques can be used to determine the number of favorable and total number.
Combinations automatically handle situations where multiple cases can occur (e.g., 3 cards with one ace: ANN, NAN & NNA).
Example: Three marbles are picked at random from a bag containing 4 red marbles and 5 white marbles. What is the probability that
a) all 3 are red b) 2 red marbles
c) 1 marble is red d) no red marbles
3+0=3
2+1=3
1+2=3
0+3=3
Notice that the number selected in the numerator is always 3.

4. Examples
1) Five cards are randomly chosen from a standard deck of 52 cards Find the probability that the following are chosen
a) all 4 aces b) No aces
c) exactly 4 diamonds d) four aces and one jack
e) at least one ace
2) Three cards are dealt. Find the probability of getting either one ace or two aces.
1 – P(no aces)

5. 3) A carton contains 200 batteries, of which 5 are defective
3) A carton contains 200 batteries, of which 5 are defective. If a random sample of 5 batteries is chosen, what is the probability that at least one is defective?
4) Thirteen cards are dealt from a well shuffled standard card deck What is the probability of getting: a) all cards from the same suit b) 7 spades, 3 hearts, and 3 clubs
c) all of the 12 face cards
e) at least one diamond
P(at least one defective) = 1 – P(no defective)
OR P(at least one defective) = P(1 def) + P(2 def) + P(3 def) + P(4 def) + P(5 def) much more complicated…
P(all any 1 suit) = P(all ♠) + P(all ♣) P(all ♥) + P(all ♦)
= 4·P(all ♠)
P(at least one diamond) = 1 – P(no diamonds)

6. Should counting always be used?
6. A die is rolled twice
a) Find the probability that 2 sixes are rolled
b) Find the probability that the face value is greater than 4 and that the second is 2.
These are easier to do using probabilities and multiplication.