Question 1090959: A car dealership has 6 red, 14 silver, and 5 black cars on the lot. Ten cars are randomly chosen to be displayed in front of the dealership. Complete parts (a) through (c) below.
Find the probability that 4 cars are red and the rest are silver.
nothing
(Round to four decimal places as needed.)
(b) Find the probability that
5
5 cars are red and
5
5 are black.
nothing
(Round to six decimal places as needed.)
(c) Find the probability that exactly 6 cars are red.
nothing
(Round to five decimal places as needed.)
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! Question: (please check if edited version is the same as original question)
A car dealership has 6 red, 14 silver, and 5 black cars on the lot. Ten cars are randomly chosen to be displayed in front of the dealership. Complete parts (a) through (c) below.
(a) Find the probability that 4 cars are red and the rest are silver.
(Round to four decimal places as needed.)
(b) Find the probability that 5 cars are red and 5 are black.
(Round to six decimal places as needed.)
(c) Find the probability that exactly 6 cars are red.
(Round to five decimal places as needed.)
Solution:
This is a hypergeometric distribution problem.
When we have A objects of one class, and B objects of another. The probability of picking a from the A class and b from the B class is given by:
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
where C(n,r) is the number of combinations of picking r objects from n, and where
C(n,r)=n!/(r!(n-r)!)
The stock to choose from is 6 red, 14 silver and 5 black, for a total of 25 cars.
(a) 4 red and 6 silver
Since the hypergeometric distribution formula above applies to only two classes, we can apply the formula in two steps:
Step 1: assume silver or red together (20 cars) as one class, and black as another. Choose 10 from the combined class and calculate probability using hypergeometric distribution.
P(10 red or silver)=C(20,10)*C(5,0)/C(25,10)
Step 2: choose 4 red and 6 black from the 10 cars chosen, again using the hypergeometric distribution.
P(4 red + 6 silver from 20)=C(6,4)*C(14,6)/C(20,10)
We then multiply the two probabilities together to get
P(4 red + 6 silver from stock of 25)
=C(20,10)*C(5,0)/C(25,10) * C(6,4)*C(14,6)/C(20,10)
=C(6,4)*C(14,6)*C(5,0)/C(25,10) after simplification
which is exactly the multinomial equivalent of the hypergeometric distribution for 4 red, 6 silver and 0 black.
Evaluating,
P(4 red+6 silver)
=C(6,4)*C(14,6)*C(5,0)/C(25,10)
=15*3003*1/3268760
=819/59432
=0.013780
(b) 5 red + 5 black
Apply the multinomial hypergeometric distribution:
P(5 red + 5 black)
=C(6,5)*C(14,0)*C(5,5)/C(25,10)
=6*1*1/3268760
=3/1634380
=0.00000183556
(c) exactly 6 cars are red
We can apply directly hypergeometric distribution by putting red as one class, and silver or black (total 14+5=19) as the other.
P(6 red + 4 (silver or black)
=C(6,6)*C(19,4)/C(25,10)
=1*3876/3268780
=0.001186
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