a bag contains 4 red balls, 3 blue balls, and 2 green balls. what is the probability of drawing a red ball…

a bag contains 4 red balls, 3 blue balls, and 2 green balls. what is the probability of drawing a red ball and then a blue ball without replacement?\na $\frac{1}{12}$\nb $\frac{2}{15}$\nc $\frac{1}{6}$\nd $\frac{2}{21}$

a bag contains 4 red balls, 3 blue balls, and 2 green balls. what is the probability of drawing a red ball and then a blue ball without replacement?\na $\frac{1}{12}$\nb $\frac{2}{15}$\nc $\frac{1}{6}$\nd $\frac{2}{21}$

Answer

Explanation:

Step1: Calculate total number of balls

Total balls = 4 + 3+ 2 = 9

Step2: Calculate probability of drawing a red ball first

Probability of red ball, $P(R)=\frac{4}{9}$

Step3: Calculate probability of drawing a blue ball second without replacement

After drawing a red ball, there are 8 balls left. Probability of blue ball, $P(B)=\frac{3}{8}$

Step4: Calculate combined probability

Since these are sequential events, we multiply the probabilities. $P(R \text{ then } B)=\frac{4}{9}\times\frac{3}{8}=\frac{12}{72}=\frac{1}{6}$

Answer:

C. $\frac{1}{6}$