Question

heather has a bag containing 4 marbles: 1 red, 1 blue, and 2 green. she draws 1 marble out of the bag, replaces it, and then draws another marble. what is p(red then blue then green)? $\frac{1}{64}$ $\frac{1}{32}$ $\frac{1}{2}$ $\frac{3}{4}$

Explanation:

Step1: Calculate probability of drawing red

The probability of drawing a red marble on the first - draw. There is 1 red marble out of 4 marbles. So $P(\text{red})=\frac{1}{4}$.

Step2: Calculate probability of drawing blue

Since the marble is replaced, the probability of drawing a blue marble on the second - draw. There is 1 blue marble out of 4 marbles. So $P(\text{blue})=\frac{1}{4}$.

Step3: Calculate probability of drawing green

Since the marble is replaced, the probability of drawing a green marble on the third - draw. There are 2 green marbles out of 4 marbles. So $P(\text{green})=\frac{2}{4}=\frac{1}{2}$.

Step4: Calculate the combined probability

Since the draws are independent events, we multiply the probabilities of each event. $P(\text{red then blue then green})=P(\text{red})\times P(\text{blue})\times P(\text{green})=\frac{1}{4}\times\frac{1}{4}\times\frac{1}{2}=\frac{1}{32}$.

Answer:

$\frac{1}{32}$