Atraffic light at a certain intersection is green 50% of the time, yellow 10% of the time, and red 40% of the time. a car approaches this intersection once each day. what is the probability that in a sequence of 10 days, four green lights, one yellow light, and five red lights are encountered? assume that each day represents an independent trial.

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Step-by-step explanation:

Probability that the car encounters a green light on the first day: .

To meet the question's conditions, the car needs to encounter another green light on the second day. Given that the colors of the light on each day are "independent," the chance that there's a green light followed by another green light will be

.

To meet the condition on the fifth day, there needs to be a yellow light. The probability that the condition is met on the first four days and on the fifth day will be .

To meet the condition on the sixth day, all prior days should meet the conditions. Besides, there needs to be a red light on the sixth day.

The question asks that the condition be met on all ten days. As a result, the probability of meeting the condition will be equal to the probability on the tenth day: .

0.17274

Step-by-step explanation:

P(G) = 50/100=0.5

P(Y) = 10/100=0.1

P(R) = 40/100=0.4

In 10 days, since they are independent,

P(G) = 0.5*0.5*0.5*0.5=0.0625

P(Y) =0.1

P(R) = 0.4*0.4*0.4*0.4*0.4=0.01024

P(G) + P(Y) + P((R)= 0.17274

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse