Question 1158727
.
(1) 6 cars are picked randomly from a garage which has 4 blacks cars and 2 white cars and 4 red cars 
find the probability of getting at MOST 1 red car.
(2) 6 cars are picked randomly from a garage which has 4 black cars and 2 white cars and 4 red cars 
find the probability of getting at LEAST 1 red car
~~~~~~~~~~~~~~~~~~~~~~~



        (1)   Find the probability of getting at   MOST  1  red car


<pre>
The total number of cars in the garage is 4 + 2 + 4 = 10.


The probability under the question is the sum of probabilities to get 0 red cars or to get 1 red car,

     P = P(0) + P(1).


The number of ways to get 0 red cars is 1: it is the case to get 4 black and 2 white cars.

The number of ways to get 1 red car of 6 picked cars is equal to the number to comprise 5 cars from (4+2) = 6 cars
of the not-red color, multiplied by 4, which is the number to select 1 red car from 4 red cars.

So, the number of ways to get 1 red car of 6 picked cars is

   C(6,5)*(C4,1) = 6*4 = 24.


The total number to get 6 cars from 10 cars is  C(10,6) = 210.


Therefore, the probability under the question (1) is  {{{(1+24)/210)}}} = {{{25/210}}} = {{{5/42}}}.    <U>ANSWER</U>
</pre>

Part &nbsp;(1) &nbsp;is solved.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2) &nbsp;&nbsp;Find the probability of getting at &nbsp;LEAST &nbsp;1 &nbsp;red car


<pre>
The probability of getting at LEAST 1 red car is the COMPLEMENT to the probability to get no one red car.

There is only one way to get no one red car: it is to get 4 black cars and 2 white cars


So, the probability to get no one red car is  P' = {{{1/C(10,6)}}} = {{{1/210}}}.    


Thus, the answer to question (2) is  P = 1 - P' = 1 - {{{1/C(10,6)}}} = 1 - {{{1/210}}} = {{{209/210}}}.
</pre>

Part &nbsp;(2) &nbsp;is solved, &nbsp;too.