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A person passes through three intersections monitored by traffic lights. The location and operation of these traffic lights is such that, to all intents and purposes, they appear to operate independently to a person travelling from one to another. The probability of a red light is $0.4$, $0.8$, and $0.5$ respectively for each of the traffic lights.

(a) Find the probability function of $X$, the number of red lights the person encounters in a single trip.

(b) Compute the mean of $X$.

(c) Assume that the waiting time for each red light is two minutes. What is the mean waiting time in one trip?


What I got so far:

$P(R1)=0.4$

$P(R2)=0.8$

$P(R3)=0.5$

$P(X=3)=0.4*0.8*0.5=0.16$

$P(X=0)=1-P(X=3)=1-0.16=0.84$

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  • $\begingroup$ Let $X$ be the number of red lights encountered. Then $X$ is a discrete random variable with possible values $0,1,2,3$. You need to find the following probabilities: $$P(X=0),\;\;P(X=1),\;\;P(X=2),\;\;P(X=3)$$ $\endgroup$
    – quasi
    Commented Aug 10, 2017 at 9:52
  • $\begingroup$ Also, there's no information about yellow lights, so you should assume that the only possibilities are red and non-red. So for example, when encountering the first light, it will be red with probability 0.4, and non-red with probability 0.6. $\endgroup$
    – quasi
    Commented Aug 10, 2017 at 9:54
  • $\begingroup$ So on a single three lane intersection, isn't the maximum number of red lights only 1 or lets say 2? I mean, I cannot encounter 3 red lights on a single intersection. Maybe if it was a straight long lane with many traffic lights then it would be more logical to me? The intersection confuses me in this problem $\endgroup$ Commented Aug 10, 2017 at 9:57
  • $\begingroup$ There are no "lanes", just three separate intersections. In other words, the vehicle will encounter $3$ separate lights. The reason for the term :"intersection" is that lights usually occur at intersections. So yes, just one long lane with $3$ separate lights, each with their own probability of being red. $\endgroup$
    – quasi
    Commented Aug 10, 2017 at 9:58
  • $\begingroup$ Oh, makes more sense now! Great. Thanks! $\endgroup$ Commented Aug 10, 2017 at 10:02

2 Answers 2

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Part (a):

For convenience of notation, let $u = 0.4,\;\;v = 0.8,\;\;w=0.5$.

\begin{align*} P(X=0)=\;&(1-u)(1-v)(1-w) = .06\\[4pt] P(X=1)=\;&u(1-v)(1-w)+(1-u)v(1-w)+(1-u)(1-v)w= .34\\[4pt] P(X=2)=\;&(1-u)vw+u(1-v)w+uv(1-w)= .44\\[4pt] P(X=3)=\;&uvw=.16\\[4pt] \end{align*} Part (b):

The mean of $X$ is $$0 \cdot P(X=0) + 1\cdot P(X=1)+2\cdot P(X=2)+3\cdot P(X=3) = 1.7\;\text{minutes}$$ Part (c):

Assume each red light last for $2$ minutes . Then on average, if a given light is red, the waiting time for that light is $1$ minute (half of $2$).

Hence, the mean waiting time will be $$u\cdot 1+v\cdot 1 + w\cdot 1 = 1.7\;\text{minutes}$$

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As we've got only 3 lights with probabilities of red light $p_1, p_2$ and $p_3$, the solution seems to be easy

$$P(X=0)=\prod\limits_{i=1}^3 (1-p_i)\\ P(X=1)=\sum\limits_{i=1}^3p_i\prod\limits_{k\neq i} (1-p_k)\\ P(X=2)=\sum\limits_{i=1}^3(1-p_i)\prod\limits_{k\neq i} p_k\\ P(X=3)=\prod\limits_{i=1}^3 p_i$$

Notice, that it is not true, that $P(X=0)=1-P(X=3)$

As You've got these probabilities, you can easily compute the mean value of $X$

Computing the mean waiting time could be calculated using yhis formula:

$$T=\sum\limits_{i=1}^3 (t_ip_i + 0(1-p_i))$$ where $t_{i}$ is waiting time on $i$'th light.

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