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. $\frac{2}{15}$
d. $\frac{1}{15}$

Explanation:

Step1: Calculate probability of grabbing a blue pen

The total number of pens is \(4 + 2=6\). The number of blue pens is 2. So the probability of grabbing a blue pen \(P(\text{blue pen})=\frac{2}{6}=\frac{1}{3}\).

Step2: Calculate probability of grabbing a yellow high - lighter

The total number of high - lighters is \(1 + 4 = 5\). The number of yellow high - lighters is 1. So the probability of grabbing a yellow high - lighter \(P(\text{yellow high - lighter})=\frac{1}{5}\).

Step3: Use the multiplication rule for independent events

Since the events of grabbing a pen and a high - lighter are independent, the probability of both events occurring is the product of their individual probabilities. \(P = P(\text{blue pen})\times P(\text{yellow high - lighter})=\frac{1}{3}\times\frac{1}{5}=\frac{1}{15}\).

Answer:

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