Francisco's game involves 3 green, 2 yellow, 4 red, and 3 black marbles. If he randomly draws three marbles from the bag, without replacement, what is the probability that he will draw a yellow, then a red, and then a black?

A. [tex]$\frac{1}{192}$[/tex]
B. [tex]$\frac{1}{72}$[/tex]
C. [tex]$\frac{3}{220}$[/tex]
D. [tex]$\frac{1}{55}$[/tex]



Answer :

To determine the probability of drawing a yellow marble first, followed by a red marble, and finally a black marble without replacement, we need to follow these steps:

1. Calculate the total number of marbles initially:
There are 3 green, 2 yellow, 4 red, and 3 black marbles.
[tex]\[ \text{Total marbles} = 3 + 2 + 4 + 3 = 12 \][/tex]

2. Probability of drawing a yellow marble first:
The probability of picking a yellow marble first is the number of yellow marbles divided by the total number of marbles.
[tex]\[ P(\text{yellow first}) = \frac{2}{12} = \frac{1}{6} \approx 0.1667 \][/tex]

3. Adjust the total number of marbles and the number of yellow marbles after drawing the first marble:
After drawing one yellow marble, 11 marbles remain in the bag. The number of yellow marbles reduces to 1.
[tex]\[ \text{Remaining marbles} = 12 - 1 = 11 \][/tex]

4. Probability of drawing a red marble second:
The probability of picking a red marble now is the number of red marbles divided by the remaining number of marbles.
[tex]\[ P(\text{red second}) = \frac{4}{11} \approx 0.3636 \][/tex]

5. Adjust the total number of marbles and the number of red marbles after drawing the second marble:
After drawing one red marble, 10 marbles remain in the bag. The number of red marbles reduces to 3.
[tex]\[ \text{Remaining marbles} = 11 - 1 = 10 \][/tex]

6. Probability of drawing a black marble third:
The probability of picking a black marble now is the number of black marbles divided by the remaining number of marbles.
[tex]\[ P(\text{black third}) = \frac{3}{10} = 0.3 \][/tex]

7. Calculate the overall probability of these sequential events happening:
The combined probability of these independent events is the product of their individual probabilities.
[tex]\[ \text{Overall probability} = P(\text{yellow first}) \times P(\text{red second}) \times P(\text{black third}) \][/tex]
[tex]\[ = \frac{1}{6} \times \frac{4}{11} \times \frac{3}{10} = \frac{1 \cdot 4 \cdot 3}{6 \cdot 11 \cdot 10} = \frac{12}{660} = \frac{1}{55} \][/tex]

Thus, the probability that Francisco will draw a yellow marble first, then a red marble, and then a black marble is:
[tex]\[ \boxed{\frac{1}{55}} \][/tex]