Question

1. erin reaches into a bag, recorded the color of the marble she picked, returned the marble and drew again. these are the marbles she has drawn so far: blue, blue, green, purple, green, blue, orange, blue, green, orange a) what is the probability that her next draw will be blue? b) what is the probability that her next draw will not be green? c) if she draws 90 more times, how many times would you expect an orange?

Explanation:

Step1: Count total draws and blue - draws

The total number of draws so far is $n = 10$. The number of times blue was drawn is $m_{blue}=4$.

Step2: Calculate probability of drawing blue

The probability $P(\text{blue})$ of drawing a blue marble on the next draw (using relative - frequency approach) is $P(\text{blue})=\frac{m_{blue}}{n}$. Substituting values, we get $P(\text{blue})=\frac{4}{10}=\frac{2}{5}$.

Step3: Count green - draws

The number of times green was drawn is $m_{green}=3$.

Step4: Calculate probability of not drawing green

The probability of drawing a green marble is $P(\text{green})=\frac{m_{green}}{n}=\frac{3}{10}$. The probability of not drawing a green marble is $P(\text{not green}) = 1 - P(\text{green})=1-\frac{3}{10}=\frac{7}{10}$.

Step5: Count orange - draws

The number of times orange was drawn is $m_{orange}=2$.

Step6: Calculate expected number of orange - draws in 90 draws

The probability of drawing an orange marble is $P(\text{orange})=\frac{m_{orange}}{n}=\frac{2}{10}=\frac{1}{5}$. If she draws $N = 90$ more times, the expected number of orange - draws $E$ is $E=P(\text{orange})\times N$. Substituting values, we get $E=\frac{1}{5}\times90 = 18$.

Answer:

a) $\frac{2}{5}$
b) $\frac{7}{10}$
c) 18