Question

At a certain gas station, \({\rm{40\% }}\)of the customers use regular gas \(\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{,35\% }}\) use plus gas\(\left( {{{\rm{A}}_{\rm{2}}}} \right)\), and \({\rm{25\% }}\) use premium\(\left( {{{\rm{A}}_{\rm{3}}}} \right)\). Of those customers using regular gas, only \({\rm{30\% }}\) fill their tanks (event \({\rm{B}}\) ). Of those customers using plus, \({\rm{60\% }}\)fill their tanks, whereas of those using premium, \({\rm{50\% }}\)fill their tanks.

a. What is the probability that the next customer will request plus gas and fill the tank\(\left( {{{\rm{A}}_{\rm{2}}} \cap {\rm{B}}} \right)\)?

b. What is the probability that the next customer fills the tank?

c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

Short answer

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right) = 0.401\\P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right) = 0.723\end{array}\)

Step-by-step solution

Introduction

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

Explanation

Denote events:

\({{\rm{A}}_{\rm{i}}}{\rm{ = \{ }}\)point i error was signaled incorrectly\({\rm{\} ,i = 1,2, \ldots ,25}}\).

The probabilities of each of the events is the same and it is \({\rm{0}}{\rm{.05}}\) (given in the exercise).

We are asked to find the probability that at least one of \({\rm{10}}\) successive points indicate a problem when in fact the process is operating correctly which is the union of events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{{\rm{10}}}}\) (at least one) \(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right)\\\mathop = \limits^{(1)} P\left( {{{\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{10}^\prime } \right)}^\prime }} \right)\\\mathop = \limits^{(2)} 1 - P\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{10}^\prime } \right)\\\mathop = \limits^{(3)} 1 - P\left( {A_1^\prime } \right) \cdot P\left( {A_2^\prime } \right) \cdot \ldots \cdot P\left( {A_{10}^\prime } \right)\\\mathop = \limits^{(4)} 1 - (1 - 0.05) \cdot (1 - 0.05) \cdot \ldots \cdot (1 - 0.05)\\ = 1 - {0.95^{10}}\\ = 0.401\end{array}\)

(1): here we use De Morgan's Law,

(2): for any event\(A,P\left( {{A^\prime }} \right) + P(A) = 1\),

(3): the events (points) are independent, so we can use the multiplication property given below,

(4): using\(A,P\left( {{A^\prime }} \right) + P(A) = 1\).

Explanation of multiplication property

Multiplication Property:

For events \({A_1},{A_2}, \ldots ,{A_n},n \in \mathbb{N}\)we say that they are mutually independent if

\(\begin{array}{l}P\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right)\\ = P\left( {{A_{{i_1}}}} \right) \cdot P\left( {{A_{{i_2}}}} \right) \cdot \ldots \cdot P\left( {{A_{{i_k}}}} \right)\end{array}\)

for every\(k \in \{ 2,3, \ldots ,n\} \), and every subset of indices\({{\rm{i}}_{\rm{1}}}{\rm{,}}{{\rm{i}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{i}}_{\rm{k}}}\).

Given \({\rm{25}}\) successive points, similarly we obtain

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right)\\\mathop = \limits^{(1)} P\left( {{{\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{25}^\prime } \right)}^\prime }} \right)\\\mathop = \limits^{(2)} 1 - P\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{25}^\prime } \right)\\\mathop = \limits^{(3)} 1 - P\left( {A_1^\prime } \right) \cdot P\left( {A_2^\prime } \right) \cdot \ldots \cdot P\left( {A_{25}^\prime } \right)\\\mathop = \limits^{(4)} 1 - (1 - 0.05) \cdot (1 - 0.05) \cdot \ldots \cdot (1 - 0.05)\\ = 1 - {0.95^{25}}\\ = 0.723.\end{array}\)

(1): here we use De Morgan's Law,

(2): for any event\(A,P\left( {{A^\prime }} \right) + P(A) = 1\),

(3): the events (points) are independent, so we can use the multiplication property given above,

(4): using\(A,P\left( {{A^\prime }} \right) + P(A) = 1\)

Therefore, the result is

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right) = 0.401\\P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right) = 0.723\end{array}\)