Problem 4: For a certain style of new automobile, the colours blue, white, black

Question

Problem 4: For a certain style of new automobile, the colours blue, white, black and green are in equal demand. Three successive orders are placed for automobiles of this style. Find the probability that: (a) one blue, one white and one green are ordered (b) two blues are ordered (c) at least one black is ordered (d) exactly two of the orders are for the same colour.

It would be assuming infinite population as equal demend and order matters as it is successive. Thus first order there are 4 options, second option 4 options and third option 4 options (4)^3=64 different permitations with relacment.

Let B=bl W=white g=green K=black

a) Differnt ways we can have BWG apper would be 3! = 6
P(BWG)=6/64

b) P(2b) we know BB?, ? has 3 options (since exaclty 2 BLUE) and we can reagrange the last coulour 3 differnt ways since we have 3 spots. 3*3=9 = 3C2*3C1

P(2b)=9/64

c) this is where I am stuck.
P(B1)=K?? ?=4*4 since we can have black occure again. - I know I could chnage the problem to the following...

We can arrange it 3 different way thus 4*4*3=48... but as stated bellow I know this is wrong.

Let Z= at least 1 black car

Zc=0 Black cars

nZc=3*3*3=27

P(Zc)=27/64
THUS P(Z)=1-27/64 = 37/64

Though I want to do it the first way also as I obviosuly have a gap in my knlowedge since I could apply the following method in a and b.

Maybe I could do the following P(1Black) + P(2Black) + P(3Black) (ALL EXACTLY)
P(1Black)= K??, ?=3 different coulours, thus 3^2 *3 differnt ways to arrange
P(1black)=3*3^2/64 = 27/64
P(2black)=KK*3 with 3 differnt ways = 9/64
P(3 Black) = KKK = 1 since we cant really rearange this = 1/64... B

P(AT LEAST 1 BLACK) = P(1bacl) + P(2black) + P(3Black) = 37/64

WHich is correct based on my first finding, but verification would be good and just like to see if my above method is correct effectivly :)

Explanation / Answer

Your first 2 answers are perfectly correct and your understanding is also correct.

Now when you say at least 1 black, then it means P(1 Black, 2 others)+ P(2 Blacks, 1 other)+ P(All Blacks)

Now P(2 blacks, 1 another) is simple= (1/64)*3!/2!= 3/64 and since 3 other colours are available, therefore (3/64)*3=9/64

and P(3 blacks)= 1/64

Now lets get to our first part--P(1 black and 2 others)---here 2 others can either be of the same colour or they can be 2 different colours. hence we will have to take their individual probabilities.

So P(1 black, 2 others of the same colour)= (1/64)*3C1*3= 9/64 (3C1--to choose 1 of any 3 colours)

and P(1 black, 2 others of different colours)= (1/64)*3C2*3!= 18/64

So the combined probability is 18/64+9/64+9/64+1/64 = 37/64, which is what you too have got.

d) I think this is very simple: choose 1 amongst the four colours in 4C1=4 ways, arrange them in 3!/2!= 3 ways, and the probability of picks = 1/64, therefore 4*3*(1/64) = 12/64 = 3/16

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