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.

To find the probability that all three balls selected at random without replacement are red, we can break down the selection process step-by-step.

Step 1: Determine the total number of balls in the box
The box contains:
- Red balls = 4
- Green balls = 4
- Blue balls = 4
Total number of balls = 4 + 4 + 4 = 12

Step 2: Calculate the probability of selecting the first red ball
Initially, there are 4 red balls out of a total of 12 balls. The probability of drawing a red ball on the first attempt is:
$$P_1=\frac{4}{12}$$

Step 3: Calculate the probability of selecting the second red ball
Since the drawing is done without replacement, there are now 11 balls remaining in the box, of which 3 are red. The probability of drawing a second red ball is:
$$P_2=\frac{3}{11}$$

Step 4: Calculate the probability of selecting the third red ball
After successfully drawing two red balls, there are 10 balls left in the box, and 2 of them are red. The probability of drawing a third red ball is:
$$P_3=\frac{2}{10}$$

Step 5: Calculate the combined probability
The overall probability of drawing three red balls in a row is the product of the individual probabilities for each step:
$$P\left(\text{All three are red}\right)=P_1\times P_2\times P_3$$
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 before multiplying:
$$P\left(\text{All three are red}\right)=\frac{1}{3}\times \frac{3}{11}\times \frac{1}{5}$$
The factor of 3 in the numerator and denominator cancels out:
$$P\left(\text{All three are red}\right)=\frac{1\times 1}{11\times 5}=\frac{1}{55}$$
Thus, the final probability that all three pulled balls are red is 1/55.