1. ## Question

A car dealer has 8 Red, 11 Gray, and 9 Blue cars in stock Ten cars are randomly chosen to be displayed in from of the dealership. Find the probability that 4 are red and 6 are blue; 3 are red, 3 are blue and 4 are gray; 5 are gray and non are blue; all 10 are gray.

2. Originally Posted by batman123
A car dealer has 8 Red, 11 Gray, and 9 Blue cars in stock Ten cars are randomly chosen to be displayed in from of the dealership. Find the probability that 4 are red and 6 are blue; 3 are red, 3 are blue and 4 are gray; 5 are gray and non are blue; all 10 are gray.
Remember that probability is favorable outcomes divided by possible outcomes.

1)4 Red and 6 Blue: There are 8 red he is chosing 4 thus there are $\displaystyle _8C_4=70$ different ways. There are 9 blue he is chosing 6 thus there are $\displaystyle _9C_6=_9C_3=84$ different ways. Thus there are a complete total of $\displaystyle 70\times 84=5880$ ways of chosing these types of cars. This is the favorable outcomes, now we find the possible outcomes. There are $\displaystyle 28$ cars altogether and he is taking 10, thus there are $\displaystyle _{28}C_{10}=13123110$ different ways. Thus the probability is the ratio thus,
$\displaystyle p=\frac{5880}{13123110}$.

Everything else is solved similarly.