There are 35 people in a group. x(x+1) of them have a blue car, 5x of them have a red car, 4 have a blue and a red car and 4x-8 do not have car. Work out the probability that a person who has a blue car, has a red car as well.

We see this question is about conditional probability. Because a person chosen at random from the group satisfies a 1st condition: they have a blue car. And after we know this, they have to satisfy a 2nd condition: they also have a red car. This probability is calculated dividing the number of people that have a red car by those with a blue car. So you need the number of people in each group...The total number of people is 35. This group is called a "set". This set is divided in smaller groups depending on the car the people own; these are subsets. Then, the sum of all the people in the subsets will be the set (35). This is x(x+1) + 5x + 4x - 8 + 4 = 35Use the quadratic formula to work out x. Ignore the negative solution because we can't have a negative number of people. You get x=3 and substitute it into the operation for each group to get the people. To make things easy, set letters as the names of the groups:(people with blue car) |A|= x(x+1) = 12(with red car) |B|= 5x= 25(with red and blue cars) |C|= 4(with no car) |D|= 4x-8 = 12Finally, how many people can have a blue car? The people in subsets A and C. So we add them: 12 + 4= 16. And how many people have both a blue and a red car? 4. Therefore, the probability is given byP(red/blue) = 4/16