1. on his way to work every day, frank passes through two intersections with traffic signals. sometimes the lights are green when he arrives at the light; sometimes they are red,
and he must stop. the probability that he must stop at the first signal is p(first) = 0.4. the probability that he must stop at the second signal is p(second) = 0.5. the probability
that he must stop at both lights is 0.3. suppose we randomly select one morning that he travels to work, and we look at the outcomes of the two lights.
a. describe the event "first, not second" in words.
b. list all of the outcomes in the sample space.
c. calculate p(first or second), and interpret your result in context.
d. calculate the probability of the event described in part (a).
e. are the events "stopping at first light" and "stopping at second light" independent? explain your answer.
f. assuming the probability of stopping at the first signal does not change from day to day, how surprising would it be for frank to have to stop at the first light five days in a row?
if he does have to stop at the first light five days in a row, would you question the model that assigns a probability of 0.5 to the first light each day?

answer: i believe the answer is e

step-by-step explanation:

step-by-step explanation:

(3rd option) - c

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

25/6 hope it : )

step-by-step explanation: