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

Question

Prepp Team · Accepted Answer

This problem asks us to find the probability of drawing three red balls consecutively from a box without replacement. Understanding the initial conditions and how the number of available balls changes after each draw is crucial for an accurate calculation.

Balls: Initial Composition in the Box

First, let's understand the contents of the box. We have balls of three different colors:

Ball Color	Quantity
Red	4
Green	4
Blue	4

The total number of balls in the box at the start is $4 (\text{red}) + 4 (\text{green}) + 4 (\text{blue}) = 12$ balls.

Probability of Drawing Three Red Balls Without Replacement

We are drawing three balls one after another without replacement. This means that once a ball is drawn, it is not put back into the box, affecting the total number of balls and the number of specific colored balls for subsequent draws. We want to find the probability that all the three balls pulled out are red.

First Ball: Probability of Red

Initially, there are 4 red balls in the box.
The total number of balls is 12.
The probability of the first ball drawn being red is calculated as the ratio of red balls to the total number of balls: $$P(\text{1st Red}) = \frac{\text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{4}{12}$$

Second Ball: Probability of Red (Given First was Red)

Since the first ball drawn was red and not replaced, the composition of balls in the box changes for the second draw:

The number of red balls remaining is $4 - 1 = 3$.
The total number of balls remaining in the box is $12 - 1 = 11$.
The probability of the second ball drawn being red, given the first was red, is: $$P(\text{2nd Red } | \text{ 1st Red}) = \frac{\text{Remaining Red Balls}}{\text{Remaining Total Balls}} = \frac{3}{11}$$

Third Ball: Probability of Red (Given First Two were Red)

Similarly, after the second red ball is drawn and not replaced, the box contents change again for the third draw:

The number of red balls remaining is $3 - 1 = 2$.
The total number of balls remaining in the box is $11 - 1 = 10$.
The probability of the third ball drawn being red, given the first two were red, is: $$P(\text{3rd Red } | \text{ 1st & 2nd Red}) = \frac{\text{Remaining Red Balls}}{\text{Remaining Total Balls}} = \frac{2}{10}$$

Overall Probability Calculation

To find the probability that all three balls pulled out are red, we multiply the probabilities of each sequential event, as these are dependent events:

$$P(\text{All three Red}) = P(\text{1st Red}) \times P(\text{2nd Red } | \text{ 1st Red}) \times P(\text{3rd Red } | \text{ 1st & 2nd Red})$$

Substituting the calculated probabilities:

$$P(\text{All three Red}) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10}$$

Let's simplify the expression by multiplying the numerators and denominators:

$$P(\text{All three Red}) = \frac{4 \times 3 \times 2}{12 \times 11 \times 10}$$ $$P(\text{All three Red}) = \frac{24}{1320}$$

Now, we simplify the fraction $\frac{24}{1320}$. We can divide both the numerator and the denominator by their greatest common divisor. We can observe that $24$ is a factor of $1320$. Alternatively, we can simplify step-by-step:

$$P(\text{All three Red}) = \frac{1}{3} \times \frac{3}{11} \times \frac{1}{5}$$

Multiply the simplified fractions:

$$P(\text{All three Red}) = \frac{1 \times 3 \times 1}{3 \times 11 \times 5}$$ $$P(\text{All three Red}) = \frac{3}{165}$$

Further simplification by dividing both the numerator and denominator by 3:

$$P(\text{All three Red}) = \frac{3 \div 3}{165 \div 3} = \frac{1}{55}$$

Therefore, the probability that all the three balls pulled out of the box are red is $\frac{1}{55}$.

Prepp Team · Answer

1/72

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1/36

Prepp Team · Answer

1/27