Question

a car distributor sells ford cars and renault cars. it has 30 cars for sale on a particular day; 18 are ford cars and 12 are renault cars. 7 of the ford cars are red and 4 of the renault cars are red. one of the 30 cars is chosen at random. what is the probability that the car chosen is a ford car or a car which is not red? anyone know this im stuck ​

Answer 1

Step-by-step explanation:

let Ford cars be F and Renault cars be R

let Red for cars be Fr and red Renault cars be Rr

Sample Space (S) = { F1,F2,F3,F4,F5,F6,F7,F8,F9,F10,F11,Fr1,Fr2,Fr3,Fr4,Fr5,Fr6,Fr7,R1,R2,R3,R4,R5,R6,R7,R8,Rr1,Rr2,Rr3,Rr4 }

let A be event in which car chosen is not red.

A = F1,F2,F3,F4,F5,F6,F7,F8,F9,F10,F11,R1,R2,R3,R4,R5,R6,R7,R8

n(A) = 19

P(A) = n(A) / n(S)

= 19/30

Answer 2

Answer:

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Step-by-step explanation:

P(Ford) = 18/30

P(not red) = 19/30

P(Ford and not red) = 11/30

P(Ford or not red) = (18+19-11)/30 = 26/30. (This subtraction eliminates double counting.)

Unless this is looking for Ford (but not "not red") and not red (but not Ford) where the "or" excludes any car that is both. 15/30