How to do probability question (v)? Why is my logic wrong? One plastic toy aeroplane is given away free in each 
packet of cornflakes. Equal numbers of red, yellow, 
green and blue aeroplanes are put into the packets.    

Henry, a quality controller employed by the cornflakes 
manufacturer, opens a number of packets chosen at random 
to check on the distribution of colours. 

Find the probability that 

(v)  the first two packets he opens have aeroplanes of different colours 


For v:
Shouldnt it be 1/4 * 1/4 too because it is equal the proportion of different colors? Or 1/4 (color A) *3/4 (different color)= 3/16? Can someone explain to me why my logic is wrong?
Ans= 1 - ((1/4)^2)4 = 3/4
 A: You can think of it this way...
It doesn't matter which color the first one is. Whether that be red, yellow, green or blue, what matters is the second color. For any first color, there are $4-1=3$ other colors. Therefore, our probability is $1\cdot \frac{3}{4} = \frac{3}{4}$, because the $1$ signifies that any color works. The $\frac{3}{4}$ signifies the other three colors that can be chosen second. Therefore, the answer is $\boxed{\frac{3}{4}}$.
-FruDe
A: The correct answer you posted (I think taken by your textbook) is self evident:
In  your recent post we saw that the probability to have the first two packets with two red aeroplanes is $(\frac{1}{4})^2$
that means that the  probability to have the first two packets with two  "same color" aeroplanes is $4 \cdot(\frac{1}{4})^2$
Thus the requested probability is the complement of the previous probability to 1, say
$1-4 \cdot(\frac{1}{4})^2=\frac{3}{4}$
EDIT: in your solution you found $\frac{1}{4} \cdot \frac{3}{4}=\frac{3}{16}$
this means, for example: "Red" and "Not Red" but you have 4 different colors to chose, so if you multiply your solution times 4 you have the correct solution.
My answer was to  lead you to the nearest solution of the previous point you posted today.
