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.

Respuesta :

Answer:

[tex]P(Gray\ and\ Blue) = \frac{1}{4}[/tex]

Step-by-step explanation:

Given

[tex]Sections = 10[/tex]

[tex]n(Gray) = 5[/tex]

[tex]n(Blue) = 5[/tex]

Required

Determine P(Gray and Blue)

Using probability formula;

[tex]P(Gray\ and\ Blue) = P(Gray) * P(Blue)[/tex]

Calculating P(Gray)

[tex]P(Gray) = \frac{n(Gray)}{Sections}[/tex]

[tex]P(Gray) = \frac{5}{10}[/tex]

[tex]P(Gray) = \frac{1}{2}[/tex]

Calculating P(Gray)

[tex]P(Blue) = \frac{n(Blue)}{Sections}[/tex]

[tex]P(Blue) = \frac{5}{10}[/tex]

[tex]P(Blue) = \frac{1}{2}[/tex]

Substitute these values on the given formula

[tex]P(Gray\ and\ Blue) = P(Gray) * P(Blue)[/tex]

[tex]P(Gray\ and\ Blue) = \frac{1}{2} * \frac{1}{2}[/tex]

[tex]P(Gray\ and\ Blue) = \frac{1}{4}[/tex]

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