Probability Problems Solved with Combinations

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!

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.

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)

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)

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.