Question 1083442: A car dealership has 7 red, 9 silver and 5 black cars on a lot. Ten cars are randomly chosen to be displayed in front of the dealership. Find probability that 4 cars are red and the rest silver. Find the probability that 5 cars are red and 5 are black. Find the probability that exactly 7 cars are red.
I'm not sure how to find the probability could you please show me the steps and help with the answers?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! There are 21 cars and 10 will be displayed. Order doesn't matter so it is 21C10 for number of ways (denominator of probability)=352716.
4 cars chosen from 7 red is 7C4. That is multiplied by 9C6, which is the number of ways to choose 6 cars from 9, which is 84
7C4=35.
Probability is 35*84/352716=2940/352716. We multiply, because each of the first can be paired with any of the second.
============
For 5 red and 5 black it is 7C5*1, since all black cars will be chosen. This has a probability of 21/352716
===========
For 7 cars to be red it is 7C7*14C3, since there are 84 ways the other 3 cars may be chosen. This gives a probability of 84/352716.
|
|
|
