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a jar contains 8 red marbles numbered 1 to 8, 10 blue marbles numbered …

Question

a jar contains 8 red marbles numbered 1 to 8, 10 blue marbles numbered 1 to 10, and 6 white marbles numbered 1 to 6. suppose a marble is drawn at random from the jar. find the probability of each given event.
write your answers as simplified fractions or whole numbers.
a. the marble is red.
the probability a red marble is drawn is

b. the marble is not red.
the probability the drawn marble is not red is

c. the marble has the number 2 written on it.
the probability a marble is drawn with the number 2 written on it is

d. the marble is blue with the number 3 written on it.
the probability the marble is blue with the number 3 written on it is

e. the marble has the number 19 written on it.
the probability a marble is drawn with the number 19 written on it is

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Explanation:

Step1: Calculate total number of marbles

Total marbles = \(8 + 10+6=24\)

Step2: Calculate probability for part A

Number of red marbles \(n(R)=8\). Probability \(P(R)=\frac{n(R)}{Total}=\frac{8}{24}=\frac{1}{3}\)

Step3: Calculate probability for part B

Probability of not - red \(P(\text{not }R)=1 - P(R)=1-\frac{1}{3}=\frac{2}{3}\)

Step4: Calculate probability for part C

There is one red marble numbered 2, one blue marble numbered 2 and one white marble numbered 2. So \(n(2)=3\). Probability \(P(2)=\frac{3}{24}=\frac{1}{8}\)

Step5: Calculate probability for part D

There is one blue marble numbered 3. So \(n(\text{blue and }3) = 1\). Probability \(P(\text{blue and }3)=\frac{1}{24}\)

Step6: Calculate probability for part E

Since the marbles are numbered from 1 - 8 (red), 1 - 10 (blue) and 1 - 6 (white), there is no marble numbered 19. So \(n(19)=0\). Probability \(P(19)=\frac{0}{24}=0\)

Answer:

A. \(\frac{1}{3}\)
B. \(\frac{2}{3}\)
C. \(\frac{1}{8}\)
D. \(\frac{1}{24}\)
E. \(0\)