Sample space probability question

[Mark53]

Homework Statement

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Driving to work, a commuter passes through a sequence of three traffic lights. At each light he either stops, denoted by s, or continues, denoted by c. Assume that the outcome c or s for each traffic light is independent of the outcome of other traffic lights.

(a) Write out the sample space Ω.
(b) If X(ω) is the number of times the commuter stops for outcome ω, calculate X for each outcome in your sample space and write out the state space S for X.
(c) Assuming that each outcome ω is equally likely, calculate the PMF fX of X, with reasoning.
(d) Assuming that stopping at a light is twice as likely as continuing through, calculate the PMF fX of X.

The Attempt at a Solution

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A)

Ω={ccc,ccs,css,sss,ssc,scc,csc,scs}

b)

x(ccc)=0
x(ccs)=1=x(csc)=x(scc)
x(css)=2=x(ssc)=x(scs)
x(sss)=3

state space ={0,1,2,3}

c)

wouldn't the pmf just be this but the questions says equally likely.

p(X=0)=1/8
p(X=1)=3/8
p(X=2)=3/8
p(X=3)=1/8

d)
not sure how to go about this part

 

[haruspex]

the questions says equally likely

No, it says the individual (triple) outcomes like ccc are equally likely. It does not say the values of X are equally likely.

d)
not sure how to go about this part

Then you do not really understand how you answered c). You were not told that c and s were equally likely.
In d), c and s are not equally likely. But you know what the two probabilities add up to.

 

[Mark53]

No, it says the individual (triple) outcomes like ccc are equally likely. It does not say the values of X are equally likely.

Then you do not really understand how you answered c). You were not told that c and s were equally likely.
In d), c and s are not equally likely. But you know what the two probabilities add up to.

read the question wrong so what i wrote in part c would be correct then.

for part d

the probability c happening would be 1/3 and the probability of s happening would be 2/3 given that it is twice as likely and must add to 1.

does this then mean P(ccc) happening would be 1/27 and so on for the rest?

 

[haruspex]

read the question wrong so what i wrote in part c would be correct then.

for part d

the probability c happening would be 1/3 and the probability of s happening would be 2/3 given that it is twice as likely and must add to 1.

does this then mean P(ccc) happening would be 1/27 and so on for the rest?

Yes.