Question

1. you have 3 black, 1 green, and 2 blue pens in a drawer. suppose you reach into the drawer without looking and choose a pen, replace it, and then choose another pen. what is the probability of randomly choosing a green pen first and then choosing a second pen that is not blue?

Explanation:

Step1: Calculate total number of pens

Total pens = 3 (black) + 1 (green) + 2 (blue)=6.

Step2: Calculate probability of choosing green pen first

Probability of choosing green pen, $P(G)=\frac{1}{6}$ since there is 1 green pen out of 6 total pens.

Step3: Calculate probability of choosing non - blue pen second

Number of non - blue pens = 3 (black)+1 (green)=4. Probability of choosing non - blue pen, $P(\text{not }B)=\frac{4}{6}$.

Step4: Calculate combined probability

Since the events are independent (because we replace the pen), we use the multiplication rule for independent events. $P = P(G)\times P(\text{not }B)$. So $P=\frac{1}{6}\times\frac{4}{6}=\frac{4}{36}=\frac{1}{9}$.

Answer:

$\frac{1}{9}$