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{1}{15}\\)

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

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

d. \\(\frac{3}{11}\\)

Explanation:

Step1: Calculate total number of pens

Total pens = 4 red + 2 blue = 6 pens

Step2: Probability of blue pen

$P(\text{blue pen}) = \frac{\text{Number of blue pens}}{\text{Total pens}} = \frac{2}{6} = \frac{1}{3}$

Step3: Calculate total number of highlighters

Total highlighters = 1 yellow + 4 green = 5 highlighters

Step4: Probability of yellow highlighter

$P(\text{yellow highlighter}) = \frac{\text{Number of yellow highlighters}}{\text{Total highlighters}} = \frac{1}{5}$

Step5: Combined probability (independent events)

$P(\text{blue pen and yellow highlighter}) = P(\text{blue pen}) \times P(\text{yellow highlighter}) = \frac{1}{3} \times \frac{1}{5} = \frac{1}{15}$

Answer:

A. $\frac{1}{15}$