Question

find each probability using multiplication

1. there are 8 green, 10 purple, 6 red and 16 blue markers in a box. a marker is chosen, replaced and then another marker is chosen. find each probability.

a) $p(\text{neither is red})$
b) $p(\text{green, then purple})$
c) $p(\text{both blue})$

Explanation:

Step1: Calculate total markers

Total markers = $8 + 10 + 6 + 16 = 40$

Step2: Solve part (a): P(neither red)

First, find non-red count: $40 - 6 = 34$. Probability one non-red: $\frac{34}{40}$. Multiply for two draws (with replacement):
$P(\text{neither red}) = \frac{34}{40} \times \frac{34}{40} = \frac{1156}{1600} = \frac{289}{400}$

Step3: Solve part (b): P(green, then purple)

Probability green first: $\frac{8}{40}$, probability purple second: $\frac{10}{40}$. Multiply:
$P(\text{green, then purple}) = \frac{8}{40} \times \frac{10}{40} = \frac{80}{1600} = \frac{1}{20}$

Step4: Solve part (c): P(both blue)

Probability blue one draw: $\frac{16}{40}$. Multiply for two draws (with replacement):
$P(\text{both blue}) = \frac{16}{40} \times \frac{16}{40} = \frac{256}{1600} = \frac{4}{25}$

Answer:

a) $\frac{289}{400}$
b) $\frac{1}{20}$
c) $\frac{4}{25}$