A motorist encounters four consecutive traffic lights, each equally likely to be red or green. Let be the number of green lights passed by the motorist before being stopped by a red light. What is the probability distribution of E?

Question

contestada

Respuesta :

warylucknow warylucknow · Answer 1 · 2020-02-14T06:27:29+01:00

Answer:

P (X = 0) = 0.50 P (X = 3) = 0.0625

P (X = 1) = 0.25 P (X = 4) = 0.0625

P (X = 2) = 0.125

Step-by-step explanation:

The random variable E is denoted as the number of green lights passed by the motorist before being stopped by a red light.

Then the random variable E can take values, E = {0, 1, 2, 3, 4}

[tex]P(X=0)=P(Red\ light)=\frac{1}{2}=0.50[/tex]

If X = 1, this implies that the motorists passes 1 green light before being stopped by the red light.

[tex]P(X=1)=P(1\ Green\ before\ Red)=(\frac{1}{2})(\frac{1}{2})=0.25[/tex]

If X = 2, this implies that the motorists passes 2 green lights before being stopped by the red light.

[tex]P(X=2)=P(2\ Green\ before\ Red)=(\frac{1}{2})^{2}(\frac{1}{2})=0.125[/tex]

If X = 3, this implies that the motorists passes 3 green lights before being stopped by the red light.

[tex]P(X=3)=P(3\ Green\ before\ Red)=(\frac{1}{2})^{3}(\frac{1}{2})=0.0625[/tex]

If X = 4, this implies that the motorists passes 4 green lights before being stopped by the red light.

[tex]P(X=4)=P(4\ Green\ before\ Red)=(\frac{1}{2})^{4}=0.0625[/tex]

The probability distribution table for the random variable E attached below.