Answer:
a
[tex]P(G = 2 , Y= 1 , R = 2) = 0.13184 [/tex]
b
[tex]P(G = R ) = 0.13919 [/tex]
Step-by-step explanation:
From the question we are told that
The probability of the traffic light being green is [tex]p_g = \frac{3}{8}[/tex]
The probability of the traffic light being red is [tex]p_r = \frac{1}{2}[/tex]
The probability of the traffic light being yellow is [tex]p_y = \frac{1}{8}[/tex]
The number of days of observation is n = 5
Generally the probability distribution function for a multinomial distribution is mathematically represented as
[tex]Pr( X_1 = x_1 \ and \ . . . \ and \ X_k =x_k) = \left \{ {{ \frac{n!}{x_1 ! , . . . x_k !} p_1^{x_1} * \cdots *p_k^{x_k} \ \ \ \ \ \ \ when \ \sum x_i = n } \atop {0 }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise \ } \right[/tex]
Generally the probability that P[G=2, Y=1, R=2]
[tex]P(G = 2 , Y= 1 , R = 2) = \frac{5!}{ 2! * 1! * 2!} * [\frac{3}{8} ]^2 * [\frac{1}{8} ]^1 * [\frac{1}{2} ] ^2[/tex]
=> [tex]P(G = 2 , Y= 1 , R = 2) = 0.13184 [/tex]
Generally the probability that P[G=R] is mathematically represented as
[tex]P(G = R ) = P( G = 2 , Y = 1 , R = 2) + P( G = 0 , Y = 5 , R = 0)+ P( G = 1 , Y = 3 , R = 1)[/tex]
Here
[tex] P( G = 0 , Y = 5 , R = 0) = \frac{5!}{ 0! * 5! * 0!} * [\frac{3}{8} ]^0 * [\frac{1}{8} ]^5 * [\frac{1}{2} ] ^0[/tex]
[tex] P( G = 0 , Y = 5 , R = 0) = 0.0000305176 [/tex]
Also
[tex] P( G = 1 , Y = 3 , R = 1)= \frac{5!}{ 1! * 3! * 1!} * [\frac{3}{8} ]^1 * [\frac{1}{8} ]^3 * [\frac{1}{2} ] ^1[/tex]
[tex] P( G = 1 , Y = 3 , R = 1)= 0.0073242 [/tex]
[tex]P(G = R ) = 0.13184 + 0.0000305176+ 0.0073242 [/tex]
[tex]P(G = R ) = 0.13919 [/tex]
