Question

unit 1 geometry quiz a (day 3)
show all work in the left column
write final answers as percentages rounded to the nearest whole number in the right column.
problem and work space

1. a box contains 6 red, 5 blue, and 9 green marbles. one marble is chosen at random.

a. what is the probability of choosing a blue marble?
\\(\frac{6}{20}\cdot\frac{5}{20}=\\)
b. what is the probability of choosing a marble that is not green?

1. a coin is flipped and a 6 - sided die is rolled.

a. what is the probability of flipping heads and rolling a 4?
b. what is the probability of flipping tails and rolling an odd number?

1. a bag contains 8 yellow and 7 purple candies. you pick one candy, do not replace it, and then pick a second candy.

a. what is the probability that both candies are yellow?
b. what is the probability that the first is purple and the second is yellow?

Explanation:

Step1: Calculate total number of marbles

Total marbles = 6 (red) + 5 (blue) + 9 (green)=20

Step2: Calculate probability of choosing blue marble (1A)

Probability = $\frac{\text{Number of blue marbles}}{\text{Total number of marbles}}=\frac{5}{20} = 0.25$
As a percentage, $0.25\times100\% = 25\%$

Step3: Calculate probability of choosing non - green marble (1B)

Non - green marbles = 6 (red)+5 (blue)=11
Probability = $\frac{\text{Number of non - green marbles}}{\text{Total number of marbles}}=\frac{11}{20}=0.55$
As a percentage, $0.55\times 100\%=55\%$

Step4: Calculate probability of coin - die events (2A)

Probability of flipping heads on a coin is $\frac{1}{2}$, probability of rolling a 4 on a 6 - sided die is $\frac{1}{6}$
Since they are independent events, probability of both = $\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}\approx 0.0833$
As a percentage, $0.0833\times100\%\approx8\%$

Step5: Calculate probability of coin - die events (2B)

Probability of flipping tails on a coin is $\frac{1}{2}$, probability of rolling an odd number (1, 3, 5) on a 6 - sided die is $\frac{3}{6}=\frac{1}{2}$
Since they are independent events, probability of both = $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4} = 0.25$
As a percentage, $0.25\times100\% = 25\%$

Step6: Calculate probability of non - replacement candy events (3A)

Probability that first candy is yellow is $\frac{8}{8 + 7}=\frac{8}{15}$
After picking one yellow candy, there are 7 yellow and 14 total candies left.
Probability that second candy is yellow given first is yellow is $\frac{7}{14}=\frac{1}{2}$
Probability that both are yellow = $\frac{8}{15}\times\frac{1}{2}=\frac{4}{15}\approx0.2667$
As a percentage, $0.2667\times100\%\approx27\%$

Step7: Calculate probability of non - replacement candy events (3B)

Probability that first candy is purple is $\frac{7}{8 + 7}=\frac{7}{15}$
After picking one purple candy, there are 8 yellow and 14 total candies left.
Probability that second candy is yellow given first is purple is $\frac{8}{14}=\frac{4}{7}$
Probability that first is purple and second is yellow = $\frac{7}{15}\times\frac{4}{7}=\frac{4}{15}\approx0.2667$
As a percentage, $0.2667\times100\%\approx27\%$

Answer:

1A. 25%
1B. 55%
2A. 8%
2B. 25%
3A. 27%
3B. 27%