Question

Suppose each day that you drive to work a traffic light that you encounter is either green with probability 3/8, red with probability 1/2, or yellow with probability 1/8, independent of the status of the light on any other day. If over the course of five days, G, Y, and R denote the number of times the light is found to be green, yellow, or red, respectively, what is the probability that P[G=2, Y=1, R=2] ? Also, what is the probability P[G=R]? [15%]. Hint: consider sequences possible in a five-day observations

Answer 1

a

`P(G = 2 ,  Y= 1 , R = 2) = 0.13184`

b

`P(G = R ) =  0.13919`

Explanation:

From the question we are told that

The probability of the traffic light being green is
`p_g = (3)/(8)`

The probability of the traffic light being red is
`p_r = (1)/(2)`

The probability of the traffic light being yellow is
`p_y = (1)/(8)`

The number of days of observation is n = 5

Generally the probability distribution function for a multinomial distribution is mathematically represented as

`Pr( X_1 = x_1 \ and \ . . .  \ and  \  X_k =x_k) =  \left \{ {{ (n!)/(x_1 ! , . . . x_k !) p_1^(x_1) * \cdots *p_k^(x_k) \ \ \ \ \ \ \  when \  \sum x_i = n } \atop {0 }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \    otherwise  \ } \right`

Generally the probability that P[G=2, Y=1, R=2]

`P(G = 2 ,  Y= 1 , R = 2) = (5!)/( 2! * 1! * 2!) *  [(3)/(8) ]^2 * [(1)/(8) ]^1 * [(1)/(2) ] ^2`

=>
`P(G = 2 ,  Y= 1 , R = 2) = 0.13184`

Generally the probability that P[G=R] is mathematically represented as

`P(G = R ) =  P( G =  2 ,  Y = 1 ,  R = 2) + P( G =  0 ,  Y = 5 ,  R = 0)+ P( G =  1 ,  Y = 3 ,  R = 1)`

Here

`P( G =  0 ,  Y = 5 ,  R = 0) =  (5!)/( 0! * 5! * 0!) *  [(3)/(8) ]^0 * [(1)/(8) ]^5 * [(1)/(2) ] ^0`

`P( G =  0 ,  Y = 5 ,  R = 0) =  0.0000305176`

Also

`P( G =  1 ,  Y = 3 ,  R = 1)=  (5!)/( 1! * 3! * 1!) *  [(3)/(8) ]^1 * [(1)/(8) ]^3 * [(1)/(2) ] ^1`

`P( G =  1 ,  Y = 3 ,  R = 1)=  0.0073242`

`P(G = R ) =  0.13184 +  0.0000305176+ 0.0073242`

`P(G = R ) =  0.13919`