Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 22E from Chapter 2 from Devore's Probability and Statistics for Engineering and Science.
Problem 22E
Chapter:
Problem:
The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is .4, the analogous probability for the second signal is .5, and the probability that he must stop at at least one of the two signals is .6. What is the probability that he must stop...
Step-by-Step Solution
Step 1
Let A be the event that the person stops at first signal.
Let B be the event that the person stops at second signal.
Given Probabilities:\[\begin{array}{l}P\left( A \right) = 0.40\\P\left( B \right) = 0.50\\P\left( {A \cup B} \right) = 0.50\end{array}\]
Step 2: (a) Stops at both signals
The probability that person stops at both signals is intersection of events A and B\[\begin{array}{l}P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\\ = 0.4 + 0.5 - 0.6\\ = 0.9 - 0.6\\ = 0.3\end{array}\]Therefore, \[P\left( {A \cap B} \right) = 0.3\]
Step 3: (b) Stops at first but not at second
The given case is intersection of event A wth compliment of B\[\begin{array}{l}P\left( {A \cap \bar B} \right) = P\left( A \right) - P\left( {A \cap B} \right)\\ = 0.40 - 0.30\\ = 0.10\end{array}\]Therefore, \[P\left( {A \cap \bar B} \right) = 0.10\]
Step 4: (c) Stops at exactly one signal
The given case is sum of events that when a person stops either only first signal or at only second signal. \[\begin{array}{l}P = \left[ {P\left( A \right) - P\left( {A \cap B} \right)} \right] + \left[ {P\left( B \right) - P\left( {A \cap B} \right)} \right]\\ = \left[ {0.4 - 0.3} \right] + \left[ {0.5 - 0.3} \right]\\ = \left[ {0.1} \right] + \left[ {0.2} \right]\\ = 0.10 + 0.20\\ = 0.30\end{array}\]Therefore, \[P = 0.30\]
Let A be the event that the person stops at first signal.
Let B be the event that the person stops at second signal.
Given Probabilities:\[\begin{array}{l}P\left( A \right) = 0.40\\P\left( B \right) = 0.50\\P\left( {A \cup B} \right) = 0.50\end{array}\]
Step 2: (a) Stops at both signals
The probability that person stops at both signals is intersection of events A and B\[\begin{array}{l}P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\\ = 0.4 + 0.5 - 0.6\\ = 0.9 - 0.6\\ = 0.3\end{array}\]Therefore, \[P\left( {A \cap B} \right) = 0.3\]
Step 3: (b) Stops at first but not at second
The given case is intersection of event A wth compliment of B\[\begin{array}{l}P\left( {A \cap \bar B} \right) = P\left( A \right) - P\left( {A \cap B} \right)\\ = 0.40 - 0.30\\ = 0.10\end{array}\]Therefore, \[P\left( {A \cap \bar B} \right) = 0.10\]
Step 4: (c) Stops at exactly one signal
The given case is sum of events that when a person stops either only first signal or at only second signal. \[\begin{array}{l}P = \left[ {P\left( A \right) - P\left( {A \cap B} \right)} \right] + \left[ {P\left( B \right) - P\left( {A \cap B} \right)} \right]\\ = \left[ {0.4 - 0.3} \right] + \left[ {0.5 - 0.3} \right]\\ = \left[ {0.1} \right] + \left[ {0.2} \right]\\ = 0.10 + 0.20\\ = 0.30\end{array}\]Therefore, \[P = 0.30\]