Question

lamont has a bag of marbles with 5 blue marbles, 4 white marbles, and 6 red marbles. find the following probabilities of lamont drawing the given marbles from the bag if the marbles are selected one a time without replacement. (give your answer as a fraction or a decimal rounded to 4 places.) a) a blue, then a red = b) a red, then a white = c) a blue, then a blue, then a blue =

Explanation:

Step1: Calculate total marbles

The total number of marbles is $5 + 4+6=15$.

Step2: Calculate probability of blue then red

The probability of drawing a blue marble first is $\frac{5}{15}$. After drawing a blue marble, there are 14 marbles left and 6 red marbles. So the probability of then drawing a red marble is $\frac{6}{14}$. The combined probability is $\frac{5}{15}\times\frac{6}{14}=\frac{30}{210}=\frac{1}{7}\approx0.1429$.

Step3: Calculate probability of red then white

The probability of drawing a red marble first is $\frac{6}{15}$. After drawing a red marble, there are 14 marbles left and 4 white marbles. So the probability of then drawing a white marble is $\frac{4}{14}$. The combined probability is $\frac{6}{15}\times\frac{4}{14}=\frac{24}{210}=\frac{4}{35}\approx0.1143$.

Step4: Calculate probability of three - blue

The probability of drawing a blue marble first is $\frac{5}{15}$. After the first blue, there are 14 marbles left and 4 blue marbles. The probability of the second blue is $\frac{4}{14}$. After the second blue, there are 13 marbles left and 3 blue marbles. The probability of the third blue is $\frac{3}{13}$. The combined probability is $\frac{5}{15}\times\frac{4}{14}\times\frac{3}{13}=\frac{60}{2730}=\frac{2}{91}\approx0.0220$.

Answer:

a) $\frac{1}{7}\approx0.1429$
b) $\frac{4}{35}\approx0.1143$
c) $\frac{2}{91}\approx0.0220$