Question

cynara has a bag of red, blue, and black pens. she randomly selects a pen, records the color, and puts it back in the bag. she does this a total of 50 times and selects 6 red pens, 10 blue pens, and 14 black pens. based on this information, what is the probability that cynara will select a blue pen, return it to the bag, and then select a red pen?

Explanation:

Step1: Find total number of pens

First, we need to find the total number of pens. The number of red pens is 6, blue is 10, and black is 14. So total pens \(= 6 + 10 + 14 = 30\).

Step2: Probability of blue pen

The probability of selecting a blue pen, \(P(\text{blue})\), is the number of blue pens divided by total pens. So \(P(\text{blue})=\frac{10}{30}=\frac{1}{3}\). Wait, no, wait. Wait, the problem is: probability of blue then red (with replacement). Wait, first, number of red pens: 6, blue:10, black:14. Total pens: \(6 + 10 + 14 = 30\). Probability of blue: \(\frac{10}{30}=\frac{1}{3}\). Probability of red: \(\frac{6}{30}=\frac{1}{5}\). Then, since the events are independent (because we return the pen), the probability of blue then red is \(P(\text{blue}) \times P(\text{red})=\frac{10}{30} \times \frac{6}{30}\). Wait, no, wait: wait, total pens: 6 red, 10 blue, 14 black. So total is \(6 + 10 + 14 = 30\). So probability of blue: \(\frac{10}{30}=\frac{1}{3}\), probability of red: \(\frac{6}{30}=\frac{1}{5}\). Then, multiplying them: \(\frac{10}{30} \times \frac{6}{30}=\frac{60}{900}=\frac{1}{15}\)? Wait, no, wait: \(\frac{10}{30} \times \frac{6}{30}=\frac{60}{900}=\frac{1}{15}\)? Wait, 106=60, 3030=900, 60/900=1/15. Wait, but let's check again. Wait, total pens: 6 + 10 +14=30. So probability of blue: 10/30=1/3. Probability of red: 6/30=1/5. Then, since we replace the pen, the two events are independent, so we multiply the probabilities. So (10/30)(6/30)= (1/3)(1/5)=1/15? Wait, no: 10/30 is 1/3, 6/30 is 1/5, multiplying gives 1/15. Wait, but let's check the answer options. One of the options is 1/15. So that's the probability.

Answer:

\(\frac{1}{15}\) (corresponding to the option with \(\frac{1}{15}\))