Four red balls, four green balls and four blue balls are put in a box. Three balls are pulled out of the box at random one after another without replacement. The probability that all the three balls are red is

The correct option is 1/55.

Let us find the probability that all three balls pulled out are red step-by-step.

First, let us calculate the total number of balls in the box:
Number of red balls = 4
Number of green balls = 4
Number of blue balls = 4
Total number of balls = 4 + 4 + 4 = 12

Three balls are drawn one after another without replacement.
Let us find the probability of drawing a red ball in each of the three successive draws:

1. For the first draw:
The number of red balls in the box is 4, and the total number of balls is 12.
The probability of drawing a red ball first is:
$P\left(R_1\right)=\frac{4}{12}$

2. For the second draw (without replacement):
Since one red ball has already been drawn, the remaining number of red balls is 3, and the total remaining number of balls is 11.
The probability of drawing a red ball second, given that the first was red, is:
$P\left(R_2\right|R_1)=\frac{3}{11}$

3. For the third draw (without replacement):
Since two red balls have already been drawn, the remaining number of red balls is 2, and the total remaining number of balls is 10.
The probability of drawing a red ball third, given that the first two were red, is:
$P\left(R_3\right|R_1\cap R_2)=\frac{2}{10}$

Now, we calculate the combined probability that all three drawn balls are red:
$P\left(\text{All three are red}\right)=P\left(R_1\right)\times P\left(R_2\right|R_1)\times P(R_3|R_1\cap R_2)$

Substituting the values:
$P\left(\text{All three are red}\right)=\frac{4}{12}\times \frac{3}{11}\times \frac{2}{10}$

Simplifying the fractions:
$\frac{4}{12}=\frac{1}{3}$
And:
$\frac{2}{10}=\frac{1}{5}$

So:
$P\left(\text{All three are red}\right)=\frac{1}{3}\times \frac{3}{11}\times \frac{1}{5}$

Canceling the common factor 3 in the numerator and denominator:
$P\left(\text{All three are red}\right)=\frac{1}{11\times 5}=\frac{1}{55}$

Alternatively, we can use combinations. The number of ways to choose 3 red balls out of 4 is:
$\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$

The total number of ways to choose any 3 balls out of 12 is:
$\left(\genfrac{}{}{0ex}{}{12}{3}\right)=\frac{12\times 11\times 10}{3\times 2\times 1}=220$

Thus, the probability is:
$\mathrm{Probability}=\frac{4}{220}=\frac{1}{55}$