Probability and Counting Methods: Tree Diagrams and Combinatorics, Exercises of Probability and Stochastic Processes

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Section 2.

Tree Diagrams

Example 2.1 Problem

For the resistors of Example 1.16, we used A to denote the event that a randomly chosen resistor is “within 50 Ω of the nominal value.” This could mean “acceptable.” We use the notation N (“not acceptable”) for the complement of A. The experiment of testing a resistor can be viewed as a two-step procedure. First we identify which machine (B 1 , B 2 , or B 3 ) produced the resistor. Second, we find out if the resistor is acceptable. Draw a tree for this sequential experiment. What is the probability of choosing a resistor from machine B 2 that is not acceptable?

Example 2.2 Problem

Traffic engineers have coordinated the timing of two traffic lights to encourage a run of green lights. In particular, the timing was designed so that with probability 0.8 a driver will find the second light to have the same color as the first. Assuming the first light is equally likely to be red or green, what is the probability P[G 2 ] that the second light is green? Also, what is P[W ], the probability that you wait for at least one of the first two lights? Lastly, what is P[G 1 |R 2 ], the conditional probability of a green first light given a red second light?

Example 2.2 Solution





1. (^5) G^1 HHH HHH HHHH
2. (^5) R 1



1. (^8) G 2 XXXXXX
2. 2 XXXXR 2



1. (^2) G 2 XXXXXX
2. 8 XXXXR 2

• G 1 G 2 0. 4
• G 1 R 2 0. 1
• R 1 G 2 0. 1
• R 1 R 2 0. 4

The tree for the two-light experi- ment is shown on the left. The prob- ability that the second light is green is P [G 2 ] = P [G 1 G 2 ] + P [R 1 G 2 ] = 0.4 + 0.1 = 0. 5. (2.1)

The event W that you wait for at least one light is the event that at least one light is red.

W = {R 1 G 2 ∪ G 1 R 2 ∪ R 1 R 2 }. (2.2)

The probability that you wait for at least one light is

P [W ] = P [R 1 G 2 ] + P [G 1 R 2 ] + P [R 1 R 2 ] = 0.1 + 0.1 + 0.4 = 0. 6. (2.3) [Continued]

Example 2.3 Problem

In the Monty Hall game, a new car is hidden behind one of three closed doors while a goat is hidden behind each of the other two doors. Your goal is to select the door that hides the car. You make a preliminary selection and then a final selection. The game proceeds as follows:

• You select a door.
• The host, Monty Hall (who knows where the car is hidden), opens one of the two doors you didn’t select to reveal a goat.
• Monty then asks you if you would like to switch your selection to the other unopened door.
• After you make your choice (either staying with your original door, or switching doors), Monty reveals the prize behind your chosen door.

To maximize your probability P[C] of winning the car, is switching to the other door either (a) a good idea, (b) a bad idea or (c) makes no difference?

Example 2.3 Solution

To solve this problem, we will consider the “switch” and “do not switch” policies separately. That is, we will construct two different tree diagrams: The first describes what happens if you switch doors while the second describes what happens if you do not switch.

First we describe what is the same no matter what policy you follow. Suppose the doors are numbered 1, 2, and 3. Let Hi denote the event that the car is hidden behind door i. Also, let’s assume you first choose door 1. (Whatever door you do choose, that door can be labeled door 1 and it would not change your probability of winning.) Now let Ri denote the event that Monty opens door i that hides a goat. If the car is behind door 1 Monty can choose to open door 2 or door 3 because both hide goats. He chooses door 2 or door 3 by flipping a fair coin. If the car is behind door 2, Monty opens door 3 and if the car is behind door 3, Monty opens door 2. Let C denote the event that you win the car and G the event that you win a goat. After Monty opens one of the doors, you decide whether to change your choice or stay with your choice of door 1. Finally, Monty opens the door of your final choice, either door 1 or the door you switched to.

[Continued]

Figure 2.

^1 /^3 H 2 ^

1 / (^3) H 1 HHH HHH (^) H 1 / (^33)



1 / (^2)  R 2 1 / 2 R^3 (^1) R 3 (^1) R 2

• G 1 / 6
• G 1 / 6
• C 1 / 3
• C 1 / 3

^1 /^3 H 2 ^

1 / (^3) H 1 HHH HHH (^) H 1 / (^33)



1 / (^2)  R 2 1 / 2 R^3 (^1) R 3 (^1) R 2

• C 1 / 6
• C 1 / 6
• G 1 / 3
• G 1 / 3

(a) Switch (b) Do Not Switch

Tree Diagrams for Monty Hall

Quiz 2.

In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed because the mobile phone may not receive the paging signal clearly. Consequently, the system will page a phone up to three times before giving up. If the results of all paging attempts are independent and a single paging attempt succeeds with probability 0.8, sketch a probability tree for this experiment and find the probability P[F ] that the phone receives the paging signal clearly.

Section 2.

Counting Methods

Example 2.4 Problem

Choose 7 cards at random from a deck of 52 different cards. Display the cards in the order in which you choose them. How many different sequences of cards are possible?

Theorem 2.

Fundamental Principle of

Counting

An experiment consists of two subexperiments. If one subexperiment has k outcomes and the other subexperiment has n outcomes, then the experiment has nk outcomes.

Example 2.

There are two subexperiments. The first subexperiment is “Flip a coin and observe either heads H or tails T .” The second subexperiment is “Roll a six-sided die and observe the number of spots.” It has six outcomes, 1 , 2 ,... , 6. The experiment, “Flip a coin and roll a die,” has 2 × 6 = 12 outcomes:

(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6).

Example 2.

Suppose there are four objects, A, B, C, and D, and we define an exper- iment in which the procedure is to choose two objects without replace- ment, arrange them in alphabetical order, and observe the result. In this case, to observe AD we could choose A first or D first or both A and D simultaneously. The possible outcomes of the experiment are AB, AC, AD, BC, BD, and CD.

Example 2.

Suppose there are four objects, A, B, C, and D, and we define an experi- ment in which the procedure is to choose two objects without replacement and observe the result. The 12 possible outcomes of the experiment are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC.