probability that a candidate comes with all $3$ pens having the same colour is $?$

Question

Question

I know this is a duplicate of this, i am still posting it because i want to know flaw in my approach.

Candidates were asked to come to an interview with $3$ pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all $3$ pens having the same colour is $?$

My Approach

Let $E=\text{chossing all three pen of same colour}=\binom{4}{3} \times 1 \times 1 \times 1=4$

i.e select which colour among red,blue,green and black and then there will be only $1$ choice

Let $F=\text{possible way of selecting 3 pen}=4 \times 4 \times 4$

As there are $4$ ways to select colour of the pen and we have to choose for $3$ pens.

Hence reqd probability$$=\frac{E}{F}$$

$$=\frac{4}{4 \times 4 \times 4}=\frac{1}{16}=0.0625$$

Where am i wrong? Please help!

Answer 1

You are wrong in denominator. It is not $4^3$ all possible ways to take 3 pens. Arangements are not important here. If you take red, blue and green it is the same as if you take blue, green and red. You can write down all possibilites.

3r

2r 1b

2r 1y

2r 1g

1r 2g

1r 2b

1r 2y

1r 1b 1g

1r 1b 1y

1r 1g 1y

0r 3g

0r 3y

0r 3b

0r 2g 1y

0r 2g 1b

0r 1g 2b

0r 1g 2y

0r 1g 1y 1b

0r 0g 2b 1y

0r 0g 1b 2y

So there is 20 possibilites.