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 the probability of drawing a red marble

The total number of marbles is \(n = 4\). The number of red marbles is \(m_{red}=1\). The probability of drawing a red marble \(P(\text{red})=\frac{1}{4}\) (using the formula \(P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)).

Step2: Calculate the probability of drawing a blue marble

Since the marble is replaced, the total number of marbles is still \(n = 4\). The number of blue marbles is \(m_{blue}=1\). The probability of drawing a blue marble \(P(\text{blue})=\frac{1}{4}\).

Step3: Calculate the probability of drawing a green marble

Since the marble is replaced, the total number of marbles is \(n = 4\). The number of green marbles is \(m_{green}=2\). The probability of drawing a green marble \(P(\text{green})=\frac{2}{4}=\frac{1}{2}\).

Step4: Calculate the combined - probability

For independent events (because the marble is replaced), \(P(A\cap B\cap C)=P(A)\times P(B)\times P(C)\). So \(P(\text{red then blue then green})=\frac{1}{4}\times\frac{1}{4}\times\frac{1}{2}\).

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Answer:

\(\frac{1}{32}\) (the second option)