Question

A car dealership has 6 red, 9 silver, and 3 black cars on the lot. Six cars are randomly chosen to be displayed in front of the dealership. Find the probability that 3 cars are red and 3 are black. 0.001077 (Round to six decimal places as needed.)

Answer 1

Answer: Hence, our required probability is 0.001077.

Explanation:

Since we have given that

Number of red cars = 6

Number of silver cars = 9

Number of black cars = 3

Total number of cars = 6+9+3=18

We need to find the probability that 3 cars are red and 3 are black.

So, the required probability is given by

`P(3R\ and\ 3B)=(^6C_3* ^3C_3)/(^(18)C_6)\\\\P(3R\ and\ 3B)=0.001077`

Hence, our required probability is 0.001077.