5B  Sample Spaces

In a coin flip there is a simple listing of what might occur - namely,

Likewise, one listing of what can happen in the throw of a die is

We call such a listing a sample space for the probability experiment. Each item listed in the sample space is called an outcome.

More precisely, a sample space for an experiment is a listing of outcomes that can occur, subject however to the following restriction:

The listing “H, T” qualifies as a sample space for a coin flip because either a head or a tail must occur, but not both, each time you flip the coin. Likewise, when you throw a die you get exactly one of the numbers 1, 2, 3, 4, 5, 6.

The restriction guarantees that any two events in a sample space are mutually exclusive, as no two of them can occur at once.

For a given experiment there may be more than one possible sample space. Another sample space for throwing a die is the listing

as obviously when you throw a die you get an even or an odd number, but not both.

example 1

The five staid members of the Oxford Mathematical Society are Adams, Bagley, Criswell, Duggens, and Enright. These gentlemen will randomly choose a committee of two to peruse an old calculus manuscript. (The word “randomly” indicates that the choice will be made in such a way that all possible outcomes are equally likely - thus they might draw names from a hat, or perhaps cards from a deck, to determine the outcome.) We write a sample space for this experiment by listing all possible committees of two:

Altogether there are C(5,2) = 10 possible outcomes. As all ten outcomes are equally likely, each has a probability of one in ten of occurring; for example, P(AB) = 1/10.

Now suppose Mr. Adams is wondering about his probability of being chosen for the committee. There are four outcomes with Adams, and as these outcomes are mutually exclusive, the probability of Adams being chosen is

Perhaps a more direct way of viewing the problem is to observe that 4 out of 10 possible committees have Adams as a member - thus his chances are 4 out of 10 of being selected, and his probability is 4/10.

For further illustration, observe also that

The preceding example demonstrates an equally likely sample space. As the name suggests, in such a sample space all the listed outcomes are equally likely. Consequently, if the sample space has N outcomes listed then each outcome has probability 1/N. If a particular event E occurs in M of those outcomes, then the probability of E is M/N.

example 2

A group of 8 singers will randomly choose 3 singers to form a trio. We compute the probability that

a)  Andy will be chosen, b)  Andy and Bob will be chosen, c)  Andy, Bob, and Cabe will be chosen, d)  Andy will not be chosen, e)  Andy or Bob will not be chosen.

The number of possible “trio committees” is C(8,3) = 56. But instead of writing them all down (a tedious task!), we will use counting methods to solve the problem.

a)  We must calculate how many possible trios contain Andy. First we place Andy in the trio, in only one way. Then we have C(7,2) = 21 ways of choosing two of the remaining seven singers to complete the trio. Thus, there are 1 · 21 = 21 possible trios containing Andy. The probability of Andy being in the trio is

b)  How many possible trios contain both Andy and Bob? First we place Andy and Bob in the trio, in only one way. Then we have six choices of singers as the third member of the trio. The number of trios with both Andy and Bob is 1 · 6 = 6, and the probability of such a trio is

c)  Since only one of the 56 possible trios has Andy, Bob, and Cabe, the probability of all three being chosen is

d)  Andy not being chosen is the negation of Andy being chosen; thus, by a) the probability he is not chosen is 1 − 3/8 = 5/8. We could solve this problem also by counting the number of trios not containing Andy. If we reject Andy, then there are only 7 singers left, so the number of possible trios without Andy is C(7,3) = 35. Therefore

e)  The event of Andy or Bob not being chosen is the negation of both Andy and Bob being chosen. Thus, by b),

example 3

A deck of 52 cards will be mixed thoroughly, and then one card will be taken at random. We compute the probability that this card is

		a)  the queen of spades,
		b)  a heart,
		c)  a red card,
		d)  a face card or a club,
		e)  a black queen or a red numerical card.

The sample space for this problem consists of a listing of all 52 cards; however, it won't be necessary to write them all down. Since all cards are equally likely to be taken, the probability of taking any one individual card is 1/52.

a)  The probability of getting the one queen of spades is 1/52.

b)  There are 13 hearts, so the probability of getting one is 13/52 = 1/4.

c)  There are 26 red cards, so the probability the card is red is 26/52 = 1/2.

d)  We must be careful about double counting. There are 12 face cards and 13 clubs, but 3 of the clubs are face cards. The number of cards that are a face card or a club is

(We subtract the 3 club face cards, as they are counted both as face cards and as clubs.) Hence the probability of getting a face card or a club is 22/52 = 11/26. We could solve this problem also with the formula

e)  There are 2 black queens and 20 red numerical cards, and there are no cards of both these descriptions. Thus there are 2 + 20 = 22 cards of one of these two types. The probability of getting such a card is 22/52 = 11/26. Alternatively, since the events “black queen” and “red numerical” are mutually exclusive, we could use the formula

example 4

We discuss experiments involving the flipping of coins. First we suppose that two coins are flipped - say a penny and a nickel to be specific. We will determine the probability of getting a) two heads, b) two tails, and c) one head and one tail.

One's first inclination might be to take as a sample space for the flipping of two coins the listing

Because one and only one of these events will occur when two coins are flipped, this listing is a valid sample space.

But the above sample space has a disadvantage in that the three listed events are not equally likely. We can understand why by looking at a more detailed sample space showing what happens individually to each of the coins. The table at the right lists four possible outcomes when a penny and nickel are flipped. We can list the four outcomes in one horizontal role as

where in each pair the first letter refers to the penny and the second to the nickel. It is clear from the symmetry involved that the four outcomes are equally likely. Since the event “two heads” occurs in only one of the outcomes, the probability of getting two heads is 1/4. Likewise, the probability of two tails is 1/4. However, the event “one head and one tail” occurs in two of the outcomes, and its probability is 2/4 = 1/2.

One can glean from the table probabilities also of other events. At least one head occurs in 3 of the 4 outcomes, so the probability of getting at least one head in flipping two coins is 3/4. Likewise the probability of at least one tail is 3/4.

The two coin problem underscores the advantage of working in equally likely sample spaces when using counting methods to determine probabilities. A common mistake among beginners is to use the sample space

in the two coin problem, and to treat all three outcomes as equally likely. Doing so leads to the false conclusion that the probability of getting two heads is 1/3. The fallacy of course is that the outcome “one head and one tail” is twice as likely as the other two outcomes in the sample space, and accordingly must be given double weight.

Next we consider the case of flipping three coins, say a penny, a nickel, and a dime. As displayed in the table there are 8 possible outcomes, all equally likely; these can be listed in abbreviated form as

where in each triple the first letter refers to the penny, the second to the nickel, and the third to the dime.

From the sample space we can determine probabilities of various events. For example,

as 3 heads occur only once, and likewise 3 tails only once. Also,

as this event occurs with three of the outcomes.

Each time we add another coin to the experiment, the size of the table doubles. For 4 coins there are 16 rows in the table, for 5 coins there are 32 rows, etc. As you would imagine, the method of writing out the sample space rapidly becomes unfeasible as the number of coins grows. For experiments involving many coins more sophisticated counting methods must be employed.

example 5

A common experiment in board games and in gambling is the throwing of two dice. In analyzing the experiment it is convenient to name the two dice - we will assume that one die is “blue” and the other “red”. The appropriate equally likely sample space is perhaps best viewed as a two-dimensional table, as displayed below. In each listed pair in the table, the first number refers to the blue die and the second to the red die - for example, the pair “4 5” indicates that 4 was thrown on the blue die and 5 on the red, as in the photo. There are 36 pairs in the table, representing 36 equally likely outcomes in the sample space.

In most games the main quantity of interest is the sum of the two numbers on the dice. In the gambling game “craps” a sum of 7 or 11 on the first throw means that the player wins the game. In many board games the sum determines how far a player can move a token around the board.

Observe that there are six outcomes in the table where the sum on the dice is 7; these are

Thus the probability that the sum equals 7 is

There are two outcomes where the sum is 11, namely “5 6” and “6 5”. Therefore the probability that the sum equals 11 is

The probability that a craps player wins the game on the first throw is

example 6

Nine puffins are lining up randomly for a picture. What is the probability that

a)  Brewster and Fenwick stand on the ends?

b)  Brewster stands next to Fenwick?

There are 9! = 362,880 ways to line up the puffins - a complete sample space would be a listing of all these ways! We can avoid this listing by using counting methods.

a)  How many ways can the puffins line up with Brewster and Fenwick on the ends? We can put Brewster and Fenwick on the ends in two ways - Brewster on the left end and Fenwick on the right, or vice versa. Then there are 7! ways of lining up the remaining puffins. Thus there are 2 · 7! ways of lining up the puffins with Brewster and Fenwick on the ends, and the probability of this event is

(Many terms in the big fraction cancel.)

b)  How many ways can the puffins line up with Brewster next to Fenwick? There are 8 ways to choose 2 slots together for Brewster and Fenwick, then there are 2! = 2 ways of putting Brewster and Fenwick in these 2 slots, and finally there are 7! ways of placing the 7 remaining puffins in the 7 remaining slots. Thus the number of ways the puffins can line up with Brewster next to Fenwick is 8 · 2 · 7!, and the probability that the two stand together is

(Everything cancels but the 2 and the 9 in the big fraction.)

example 7

The mailman is approaching the mailbox of Curly, Mo, and Larry with 2 letters. If we assume that all 3 men are equally likely to receive letters, what is the probability that

		a)  Curly gets both letters?
		b)  Mo gets at least one letter?
		c)  two different people get letters?
		d)  Curly gets more letters than Larry?
		e)  Larry and Mo each get a letter?

This problem is a little like the two coin problem, except that the coins now have three sides. For each of the two letters there are three equally likely possibilities - it can go to Curly, Mo, or Larry. We name the two letters “first letter” and “second letter”, and list all the possible outcomes for two letters in the table at the right. The nine outcomes can be listed in one line as

It is a simple matter to count outcomes in the sample space to answer the questions.

a)  The probability that Curly gets both letters is 1/9, as this event occurs only with the first outcome.

b)  Mo gets at least one letter in 5 outcomes; the corresponding probability is 5/9.

c)  In 6 outcomes two different people get letters, making the probability of this event 6/9 = 2/3.

d)  The probability that Curly gets more letters then Larry is 3/9 = 1/3, as this event occurs in 3 of the outcomes.

e)  Larry can get the first letter and Mo the second, or vice versa; the probability each gets a letter is 2/9.

EXERCISES 5B

1. In the flipping of two coins, say a penny and a nickel, what is the probability that
   1. the two coins come up alike (i.e., both heads or both tails)?
   2. the two coins come up different?
   
   
2. In the flipping of a penny, a nickel, and a dime, what is the probability that
   1. all three coins come up alike?
   2. at least one head is showing?
   3. at least two heads are showing?
   4. the penny is different from both the nickel and the dime?
   5. no two coins come up alike?
   6. at least two coins come up alike?
   
   
3. Farmer Yee's four horses are named Amber, Beauty, Concho, and Dusty. He will choose two horses at random to sell to Farmer Kealoha. What is the probability that

   		a.  Amber will be sold?
   		b.  Beauty and Concho will be sold?
   		c.  Beauty or Concho will be sold?
   		d.  Dusty will be separated from Amber?
   		e.  Beauty and Concho will wind up together?
   		f.  Amber or Beauty or Concho will be sold?

   
   
4. Seven ducks are swimming on a pond. Four of these ducks will be chosen at random and donated to a children's zoo. Calculate the probability that

   a.  Mildred will be chosen.
   b.  Mildred and Norman will be chosen.
   c.  Mildred, Norman, and Opal will be chosen.
   d.  Mildred, Norman, Opal, and Quincy will be chosen.
   e.  Mildred will not be chosen.
   f.  Mildred or Norman will not be chosen.
   g.  Mildred or Norman or Opal or Quincy will be chosen.
   h.  Mildred, Norman, and Opal will wind up together.

   
   
5. Cathy's dog grabbed one of her playing cards at random and ate it. Determine the probability the dog ate

   			a.  the jack of clubs.
   			b.  a spade or a heart.
   			c.  a diamond or a king.
   			d.  a red face card.
   			e.  a heart or a numerical card.
   			f.  a black and even numerical card.
   			g.  a three or a six or a queen.
   			h.  a red card or a black card.
   			i.  a black ten or a red face card.
   			j.  a green ace.

   
   
6. Cathy's dog next played with Cathy's two dice, rolling them around on the floor until they stopped in a random position. Determine the probability the dice landed so that

   a.  the sum of the numbers is 6,		e.  the numbers on the dice are the same,
   b.  the sum is 10,		f.  the sum is an even number,
   c.  the sum is less than 4,		g.  the sum is 2 or 12.
   d.  the sum is 10 or larger,		h.  the blue die shows a larger number than the red die.

   
   
7. Doug and Helen plan to have two children. If we assume that girls and boys are equally likely (which is very close to being true), what is the probability they will have
   1. 2 boys?
   2. 2 girls?
   3. a boy and a girl?
   
   
8. Assume now that Doug and Helen, despite their intentions, will in fact have 3 children. What then is the probability they will have
   1. 3 girls?
   2. 2 girls and 1 boy?
   3. at least 1 girl?
   4. 3 children of the same gender?
   5. at least 2 boys, with the oldest child a boy?
   
   
9. The four bridesmaids - Ann, Beth, Cheryl, and Dot - are lining up randomly behind the bride for a picture. Calculate the probability that

   Ann will be on an end, Ann and Beth will be on the ends, Cheryl will be next to Dot, Cheryl will be between Beth and Dot, the four line up alphabetically from the left, Ann will be next to Beth, and Cheryl next to Dot.

   
   
10. Mr. Xavier, Ms. Yang, and Mr. Zito teach the three sections of History 101. Bruce and Julie will each choose a section at random. What is the probability that
    1. Julie will choose Ms. Yang?
    2. Bruce and Julie wind up in the same section?
    3. Bruce will choose a male teacher, and Julie a female?
    4. one of the two will choose a male teacher, and the other a female?
    5. both will choose a male teacher?
    
    
11. Aaron, Ben, Cathy, and Dina enter a room where there are two chairs, and a couch that seats two people. If people take seats randomly, what is the probability that

    		a.  two females sit on the couch?
    		b.  Aaron sits on the couch?
    		c.  Aaron sits on the couch with a female?
    		d.  at least one female sits on the couch?