Three car brands A, B, and C have all the market share in a certain city. Brand A has 20% of the market share, brand B has 30%, and brand C has 50%. The probability that a brand A car needs a major repair during the first year of purchase is 0.05, the probability that a brand B car needs a major repair during the first year of purchase is 0.10, and the probability that a brand C car needs a major repair during the first year of purchase is 0.15. a. What is the probability that a randomly selected car in the city needs a major repair during its first year of purchase? b. If a car in the city needs a major repair during its first year of purchase, what is the probability that it is a brand A car?

a.) 0.115

b.) 0.087

Step-by-step explanation:

Given,

P(A) = 20% = 0.20

P(B) = 30% = 0.30

P(C) = 50% = 0.50

Let M - major repairing during first year

P(M/A ) = 0.05

P(M/B) = 0.10

P(M/C) = 0.15

a.)

P(M) = P(M∩A) + P(M∩B) + P(M∩C)

       = P(M/A).P(A) + P(M/B).P(B) + P(M/C).P(C)

      = (0.05)(0.20) + (0.30)(0.10) + (0.15)(0.50)

     = 0.01 + 0.03 + 0.075

     = 0.115

⇒P(M) = 0.115

b.)

P(A/M) = P(A∩M)/ P(M) = P(M/A).P(A)/P(M)

                                     = (0.05)(0.20) / 0.115

                                     = 0.01 / 0.115 = 0.087

⇒P(A/M) = 0.087

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.