Question

jeff has 4 red pens and 2 blue pens in his backpack. he also has 1 yellow highlighter and 4 green highlighters in his backpack. if he reaches into his backpack and grabs one pen and one highlighter without looking, what is the probability that he will grab a blue pen and a yellow highlighter?
a. \\(\frac{3}{11}\\)

b. \\(\frac{4}{15}\\)

c. \\(\frac{1}{15}\\)

d. \\(\frac{2}{15}\\)

Explanation:

Step1: Find total number of pens

Total pens = red pens + blue pens = \( 4 + 2 = 6 \)

Step2: Probability of blue pen

Probability of blue pen (\( P_{blue} \)) = \( \frac{\text{number of blue pens}}{\text{total number of pens}} = \frac{2}{6} = \frac{1}{3} \)

Step3: Find total number of highlighters

Total highlighters = yellow highlighters + green highlighters = \( 1 + 4 = 5 \)

Step4: Probability of yellow highlighter

Probability of yellow highlighter (\( P_{yellow} \)) = \( \frac{\text{number of yellow highlighters}}{\text{total number of highlighters}} = \frac{1}{5} \)

Step5: Probability of both events (independent)

Since grabbing a pen and a highlighter are independent events, the combined probability is \( P_{blue} \times P_{yellow} = \frac{1}{3} \times \frac{1}{5} = \frac{1}{15} \)

Answer:

C. \( \frac{1}{15} \)