16.4 Probability Problems Solved with Combinations 16.4 Warm Up Three cards are drawn from a well-shuffled standard deck of 52 cards, one after the other and without replacement. 1. Find the probability of drawing a) all clubs 13 12 11 11 b) no clubs 39 38 37 703 52 51 50 52 51 50 850 1700 c) exactly one club (hint: the club can occur on the 1st, 2nd or 3rd drawing) 13 39 38 741 3 52 51 50 1700 2. Evaluate: a) c) 13 C3 52 C3 11 850 13 C139 C2 52 C3 b) 39 C3 52 C3 703 1700 741 1700 3. Compare your answers to exercises 1 and 2. What do you notice? They are the same! 4. Write and evaluate an expression using combinations to find the probability of getting exactly 2 clubs. 13 C2 39 C1 117 52 C3 850 16.4: 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: P (E ) n(E ) # favorable n(S ) # total 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 1 4 C3 5 C0 a) all 3 are red b) 2 red marbles 4 C2 5 C1 5 9 C3 c) 1 marble is red 9 C3 21 3+0=3 10 4 C15 C2 21 9 C3 d) no red marbles 4 C0 5 C3 9 C3 1+2=3 Notice that the number selected in the numerator is always 3. 14 2+1=3 5 42 0+3=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 48 C14 C4 b) No aces 35,673 1 48 C5 52 C5 52 C5 54,145 c) exactly 4 diamonds 13 C4 39 C1 52 C5 54,145 d) four aces and one jack 4 C4 4 C1 143 13,328 52 C5 1 649,740 e) at least one ace 1 – P(no aces) 1 48 C5 52 C5 18,472 54,145 2) Three cards are dealt. Find the probability of getting either one ace or two aces. 48 4 C148 C2 4 C2 48 C1 52 C3 52 C3 221 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? P(at least one defective) = 1 – P(no defective) 1 195 C5 200 C5 1 0.12 0.88 OR P(at least one defective) = P(1 def) + P(2 def) + P(3 def) + P(4 def) + P(5 def) much more complicated… 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 P(all any 1 suit) = P(all ♠) + P(all ♣) + P(all ♥) + P(all ♦) 13 C13 4 = 4·P(all ♠) 52 C13 13 C7 13 C3 13 C3 52 C13 2.21X10 4 6.30 X 10 12 c) all of the 12 face cards 12 C12 40 C1 52 C13 6.20 X 1011 e) at least one diamond P(at least one diamond) = 1 – P(no diamonds) 1 39 C5 52 C5 1 0.22 0.78 Should counting always be used? 6. A die is rolled twice a) Find the probability that 2 sixes are rolled 2 1 1 6 36 b) Find the probability that the face value is greater than 4 and that the second is 2. 2 1 1 6 6 18 These are easier to do using probabilities and multiplication.