fraciscos game involves 3 green, 2 yellow, 4 red, and 3 black marbles. if he randomly draws three marbles…

fraciscos 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 yellow, and then red, and then black?\n\\(\\frac{1}{192}\\)\n\\(\\frac{1}{72}\\)\n\\(\\frac{3}{220}\\)\n\\(\\frac{1}{55}\\)

fraciscos 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 yellow, and then red, and then black?\n\\(\\frac{1}{192}\\)\n\\(\\frac{1}{72}\\)\n\\(\\frac{3}{220}\\)\n\\(\\frac{1}{55}\\)

Answer

Explanation:

Step1: Calculate total marbles

Total marbles = 3 + 2+4 + 3=12

Step2: Probability of drawing yellow first

Probability of drawing yellow first, $P(Y)=\frac{2}{12}=\frac{1}{6}$

Step3: Probability of drawing red second

After drawing a yellow marble, there are 11 marbles left. Probability of drawing red second, $P(R)=\frac{4}{11}$

Step4: Probability of drawing black third

After drawing a yellow and a red marble, there are 10 marbles left. Probability of drawing black third, $P(B)=\frac{3}{10}$

Step5: Calculate combined probability

Since these are dependent - events, the probability of drawing yellow, then red, then black is $P = P(Y)\times P(R)\times P(B)=\frac{1}{6}\times\frac{4}{11}\times\frac{3}{10}=\frac{1\times4\times3}{6\times11\times10}=\frac{12}{660}=\frac{1}{55}$

Answer:

$\frac{1}{55}$