jeff has 4 red pens and 2 blue pens in his backpack. he also has 1 yellow highlighter and 4 green…

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?\na. $\frac{3}{11}$\nb. $\frac{4}{15}$\nc. $\frac{2}{15}$\nd. $\frac{1}{15}$

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?\na. $\frac{3}{11}$\nb. $\frac{4}{15}$\nc. $\frac{2}{15}$\nd. $\frac{1}{15}$

Answer

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})