Question

A spinner has 10 equally sized sections, 5 of which are gray and 5 of which are blue. The spinner is spun twice. What is the probability that the first spin lands on gray and the second spin lands on blue? Write your answer as a fraction in the simplest form.

Answer 1

Answer:

$P(Gray\ and\ Blue) = \frac{1}{4}$

Step-by-step explanation:

Given

$Sections = 10$

$n(Gray) = 5$

$n(Blue) = 5$

Required

Determine P(Gray and Blue)

Using probability formula;

$P(Gray\ and\ Blue) = P(Gray) * P(Blue)$

Calculating P(Gray)

$P(Gray) = \frac{n(Gray)}{Sections}$

$P(Gray) = \frac{5}{10}$

$P(Gray) = \frac{1}{2}$

Calculating P(Gray)

$P(Blue) = \frac{n(Blue)}{Sections}$

$P(Blue) = \frac{5}{10}$

$P(Blue) = \frac{1}{2}$

Substitute these values on the given formula

$P(Gray\ and\ Blue) = P(Gray) * P(Blue)$

$P(Gray\ and\ Blue) = \frac{1}{2} * \frac{1}{2}$

$P(Gray\ and\ Blue) = \frac{1}{4}$