Answer to Question #340124 in Statistics and Probability for Maneo

Question #340124

A person driving to work every day on a route with four traffic lights believes the following to be suitable probabilities for the number of red lights encountered on a trip. Let the random variable 𝑋 be the number of red lights encountered.

Let

A be the event that no red light is encountered with P(A) = 0.05,

B be the event that one red light is encountered with P(B) = 0.25,

C be the event that two red lights are encountered with P(C) = 0.36,

D be the event that three red lights are encountered with P(D) = 0.26,

and E be the event that four red lights are encountered with P(E) = 0.08.


1) Does these probabilities satisfy the axioms of probability?

2) What is the probability of encountering at least one red traffic light on a trip?

3) What is the probability of encountering more than two red traffic lights on a trip?

4) What is the probability of encountering at the most two red traffic lights on a trip?




1
Expert's answer
2022-05-17T09:56:05-0400

1)


0<P(A)<1,0<P(B)<1,0<P(C)<1,0<P(A)<1, 0<P(B)<1,0<P(C)<1,

0<P(D)<1,0<P(E)<10<P(D)<1,0<P(E)<1

P(A)+P(B)+P(C)+P(D)+P(E)P(A)+P(B)+P(C)+P(D)+P(E)

=0.05+0.25+0.36+0.26+0.08=1=0.05+0.25+0.36+0.26+0.08=1

These probabilities satisfy the axioms of probability.


2)


P(R1)=1P(A)=10.05=0.95P(R\ge1)=1-P(A)=1-0.05=0.95

3)


P(R>2)=P(D)+P(E)=0.26+0.08=0.34P(R>2)=P(D)+P(E)=0.26+0.08=0.34

4)


P(R2)=1(P(D)+P(E))P(R\le2)=1-(P(D)+P(E))




=10.34=0.66=1-0.34=0.66

Or


P(R2)=P(A)+P(B)+P(C)P(R\le2)=P(A)+P(B)+P(C)

=0.05+0.25+0.36=0.66=0.05+0.25+0.36=0.66


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