Consider randomly selecting a single individual and having that person test drive 3 different vehicles. define events a1, a2, and a3 by a1 = likes vehicle #1 a2 = likes vehicle #2 a3 = likes vehicle #3. suppose that p(a1) = 0.55, p(a2) = 0.65, p(a3) = 0.70, p(a1 ∪ a2) = 0.80, p(a2 ∩ a3) = 0.50, and p(a1 ∪ a2 ∪ a3) = 0.84. (a) what is the probability that the individual likes both vehicle #1 and vehicle #2? (b) determine p(a2 | a3). (round your answer to four decimal places.) p(a2 | a3) =

a. Use the inclusion/exclusion principle.

b. By definition of conditional probability,

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 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.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

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

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .