The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.4, the analogous probability for the second signal is 0.5, and the probability that he must stop at at least one of the two signals is 0.6. What is the probability that he must stop a) At both signals? b) At the first signal but not the second one? c) At exactly one signal?

Assumption

Let A be the event that the person stops at first signal.

Let B be the event that the person stops at second signal.

Given

Probabilities:

P(A)=0.40

P(B)=0.50

P(A∪B)=0.50

(a) Stops at both signals

The probability that person stops at both signals is intersection of events A and B

P(A∩B) = P(A)+P(B)−P(A∪B)

           = 0.4+0.5−0.6=0.9−0.6

           =0.3

(b) Stops at first but not at second

The given case is intersection of event A wth compliment of B

P(A∩¯B) = P(A)−P(A∩B)

              =0.40−0.30

               =0.10

(c) Stops at exactly one signal

The given case is sum of events that when a person stops either only first signal or at only second signal.

P = [P(A)−P(A∩B)]+[P(B)−P(A∩B)]

  = [0.4−0.3]+[0.5−0.3]

  = [0.1]+[0.2]

  = 0.10+0.20

  =0.30

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.