Probability We will discuss different aspects of probability, from its definition to the various rules associated with probability. From independent events to disjoint events to events with replacement to events without replacement. The world of probability is vast and very useful. For the most part on the PSATs you will multiply or add probabilities depending on the context of the problem. Definition Chance Experiment - A chance experiment is any activity or situation in which there is uncertainty about which of two or more possible outcomes will result. Definition Sample Space - The collection of all possible outcomes of a chance experiment is the sample space for the experiment. Definition Probability The probability of an event E, denoted by
, is the ratio of the
number of outcomes favorable to E to the total number of outcomes in the sample space:
Important: This method for calculating probabilities is appropriate only when the outcomes of an experiment are equally likely. Ex A balanced coin is to be tossed. What is the sample space? What is the probability of each outcome in the sample space?
S = { H ,T } 1 1 , P (T ) = 2 2 Ex On some football teams the honor of calling the toss at the beginning of a football game is determines by random selection. Suppose this week a member of the offensive team will call the toss. There are 5 interior linemen on the 11player offensive team. If we define event L as the event that a lineman is selected to call the toss, 5 of the 11 possible outcomes are included in L. What is the probability that a lineman will be selected? P (H ) =
P (L) =
5 11
Ex One card is to be drawn from a deck of 52 cards and the denomination and suit recorded. (a) What is the probability that a spade is drawn?
P (Spade) =
13 1 = 52 4
(b) What is the probability that a queen is drawn?
P (Queen ) =
4 1 = 52 13
Ex A box contains five red and three green balls. One ball is to be randomly selected from the box and its color recorded. Find the sample space for this experiment and the probability of each outcome in the sample space.
S = {green, red} P ( red ) =
5
8
P (green ) =
3 8
Definition Disjoint - Two events that have no common outcomes are said to be disjoint or mutually exclusive Definition Independent - Two events E and F are said to be independent if the chance of E having occurred is unaffected by our knowledge that F has occurred. Basic Properties of Probability 1. For any event E,
.
2. If S is the sample space for an experiment, 3. If two events E and F are disjoint, then 4. The events E and F are independent iff 5. For any event E,
and
. Therefore
. .
Ex A balanced coin is to be tossed twice and the result of each toss is recorded. Find the probability of each outcome in the sample space. The coin tosses are independent since tossing a head or tail the first time has no effect on what we will toss the second time. We will use the 4th property of probability. S = { HH , HT ,TH ,TT }
P ( HH ) =
1 1 1 1 1 1 1 1 1 1 1 1 ⋅ = , P ( HT ) = ⋅ = , P (TH ) = ⋅ = , P (TT ) = ⋅ = 2 2 4 2 2 4 2 2 4 2 2 4
Ex To access a certain application in the ABC Companyʼs computer network, a password consisting of three digits followed by two letters must be correctly entered. If an unauthorized person knows how the password is constructed, what is the probability of correctly guessing the password on the first try? These events are independent. We will use the 4th property of probability.
P (Correct Password ) =
1 1 1 1 1 ⋅ ⋅ ⋅ ⋅ 10 10 10 26 26
Ex From a large volume of pea pods, a plant experimenter broke open 20 pods and counted the number of peas in each pod. The results were as follows: 5 pods contained 4 peas each, 9 pods contained 5 peas each, 4 pods contained 6 peas each, and 2 pods contained 7 peas each (a) What is the probability that a pea pod will contain fewer than 6 peas? These events are disjoint. A pea pod cannot contain 4 peas and 5 peas at the same time. We will use the 3rd property of probability.
5 9 14 7 + = = 20 20 20 10 (b) What is the probability that a pea pods will contain 6 or more peas? There are two ways to approach this problem. P ( peas < 6 ) = P ( pea = 4 ) + P ( peas = 5 ) =
Method 1:
P ( peas ≥ 6 ) = P ( pea = 6 ) + P ( peas = 7 ) = Method 2: Use the 5th property of probability.
P ( peas ≥ 6 ) = 1 − P ( pea < 6 ) = 1 −
3 7 = 10 10
4 2 6 3 + = = 20 20 20 10
Ex Suppose that in a standard deck of cards, each ace is assigned the value 0, each face card is assigned the value 1, and the remaining cards are assigned the number on the card. One card is to be selected and the value of that card noted. (a) Find the sample space for this experiment and the probability of each outcome in that sample space.
S = {0,1, 2, 3, 4, 5, 6, 7, 8, 9} P (Card = 0 ) = P (Card = 2 ) = ... = P (Card = 9 ) = P (Card = 1) =
4 1 = 52 13
12 3 = 52 13
(b) Find the probability that a card with a value of at least 7 will be drawn.
P (Card ≥ 7 ) = P (Card = 7 ) + P (Card = 8 ) + P (Card = 9 ) =
1 1 1 3 + + = 13 13 13 13
(c) Find the probability that a card with values less than 3 will be drawn.
P (Card < 3) = P (Card = 0 ) + P (Card = 1) + P (Card = 2 ) =
1 3 1 5 + + = 13 13 13 13
Examples with Replacement Ex Three slips of paper -- each with the number 1, 2, or 3 written on it -- are placed in a box. A slip of paper is randomly selected, the number on it is recorded, and the slip is then returned to the box. Finally, a second slip is randomly selected and the number on it recorded. (a) Write a sample space for this experiment.
S = {(1,1) , (1, 2 ) , (1, 3) , ( 2,1) , ( 2, 2 ) , ( 2, 3) , ( 3,1) , ( 3, 2 ) , ( 3, 3)} (b) What is the probability that the sum of the two numbers drawn is 4?
P (Sum = 4 ) =
3 1 = 9 3
(c) What is the probability that the first number drawn is a 2 or that the sum of the numbers drawn is a 3?
4 9 Note : These events are NOT disjoint (i.e. you can have a first number drawn that is 2 AND the sum still be 3), so property 3 of probability does not apply. You must count the possible outcomes to solve this problem. P ( First number drawn = 2 or Sum = 3) =
(d) What is the probability that the sum of the two numbers drawn is 8?
P (Sum = 8 ) =
0 = 0 It isnʼt possible. 9
Ex Three cards are to be randomly selected, in succession, with replacement, from a deck of 52 cards. What is the probability that the first card drawn will be a heart, the second card a hear, and the third a spade?
P (1st Card = Heart and 2 nd Card = Heart and 3rd Card = Spade) =
13 13 13 1 ⋅ ⋅ = 52 52 52 64
Note : These events ARE independent, so we can use property 4 of probability. The fact that the cards are replaced means the events are independent. Examples without Replacement Ex Suppose two calculators are to be randomly selected, in succession, without replacement, from a box that contains four defective and nine good calculators; after each selection the calculator is checked to see whether it is good or defective. What is the probability that the first calculator is good and the second calculator selected is defective. Note : These events are NOT independent (i.e. choosing a defective or good calculator affects what you will choose next), so we cannot use property 4 of probability. The fact that the calculators are not replaced makes the events dependent.
P (1st Calc = Good and 2 nd Calc = Defective) =
9 4 3 ⋅ = 13 12 13
Ex Two cards are to be randomly selected, in succession, without replacement, from a deck of 52 cards. What is the probability that the first card drawn will be a heart and the second card drawn will be a spade?
P (1st Card = Heart and 2 nd Card = Spade) =
13 13 13 ⋅ = 52 51 204
Ex From a box containing five red balls and three green balls, four balls are to be randomly selected, in succession, without replacement. What is the probability that the first ball will be red, the second ball green, and the last two red?
P (1st Ball = Red and 2 nd Ball = Green and 3rd Ball = Red and 4 th Ball = Red ) =
5 3 4 3 3 ⋅ ⋅ ⋅ = 8 7 6 5 28
