91Ó°ÊÓ

A key ring contains four office keys that are identical in appearance, but only one will open your office door. Suppose you randomly select one key and try it. If it does not fit, you randomly select one of the three remaining keys. If it does not fit, you randomly select one of the last two. Each different sequence that could occur in selecting the keys represents one of a set of equiprobable simple events. a. List the simple events in \(S\) and assign probabilities to the simple events. b. Let \(x\) equal the number of keys that you try before you find the one that opens the door \((x=1,2,3,4)\). Then assign the appropriate value of \(x\) to each simple event. c. Calculate the values of \(p(x)\) and display them in a table. d. Construct a probability histogram for \(p(x)\).

Step by step solution

01

Listing Simple Events and Assigning Probabilities

The possible sequences of selecting the keys are as follows: - Selecting the correct key on the first try: A (Probability: 1/4) - Selecting the correct key on the second try: BA (B is wrong, A is correct; Probability: 3/4 * 1/3 = 1/4) - Selecting the correct key on the third try: BCA (BC is wrong, A is correct; Probability: 3/4 * 2/3 * 1/2 = 1/4) - Selecting the correct key on the fourth try: BCD (BCD is wrong, A is correct; Probability: 3/4 * 2/3 * 1/2 * 1 = 1/4)

02

Assign Value of x to Each Simple Event

Now we will assign value of x, which represents the number of tries before finding the key that opens the door, to each simple event: - A: x=1 - BA: x=2 - BCA: x=3 - BCD: x=4

03

Calculate the Values of p(x)

Now we will calculate the probability for each value of x: - p(x=1): P(A) = 1/4 - p(x=2): P(BA) = 1/4 - p(x=3): P(BCA) = 1/4 - p(x=4): P(BCD) = 1/4 Now, we will display these probabilities in a table: | x | p(x) | |---|------| | 1 | 1/4 | | 2 | 1/4 | | 3 | 1/4 | | 4 | 1/4 |

04

Construct Probability Histogram for p(x)

Lastly, we need to construct the probability histogram. In the histogram, x is represented on the horizontal axis and p(x) is represented on the vertical axis. Here is the probability histogram for the given scenario: |x | 1 | 2 | 3 | 4 | |--|-----|-----|-----|-----| |p(x)|1/4 |1/4 |1/4 |1/4 | For this histogram, each bar is equally likely, since the probability of finding the correct key in 1, 2, 3, and 4 tries are all equal to 1/4.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!