a spinner contains four equally shaded areas, as shown below. louise spins the spinner twice. which…

a spinner contains four equally shaded areas, as shown below. louise spins the spinner twice. which probabilities are correct? check all that apply. p(both green) = 1/16 p(both blue) = 1/4 p(first green and then blue) = 1/6 p(first orange and then blue) = 1/8 p(first orange and then green) = 1/4

a spinner contains four equally shaded areas, as shown below. louise spins the spinner twice. which probabilities are correct? check all that apply. p(both green) = 1/16 p(both blue) = 1/4 p(first green and then blue) = 1/6 p(first orange and then blue) = 1/8 p(first orange and then green) = 1/4

Answer

Explanation:

Step1: Calculate probability of single - spin

Since there are 4 equally - shaded areas, the probability of landing on any one color in a single spin is $P(\text{any color})=\frac{1}{4}$.

Step2: Use multiplication rule for independent events

For two independent spins, if $A$ and $B$ are the events of the first and second spins respectively, $P(A\cap B)=P(A)\times P(B)$.

For $P(\text{both green})$:

$P(\text{green on first spin})=\frac{1}{4}$ and $P(\text{green on second spin})=\frac{1}{4}$, so $P(\text{both green})=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$.

For $P(\text{both blue})$:

$P(\text{blue on first spin})=\frac{1}{4}$ and $P(\text{blue on second spin})=\frac{1}{4}$, so $P(\text{both blue})=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\neq\frac{1}{4}$.

For $P(\text{first green and then blue})$:

$P(\text{green on first spin})=\frac{1}{4}$ and $P(\text{blue on second spin})=\frac{1}{4}$, so $P(\text{first green and then blue})=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\neq\frac{1}{6}$.

For $P(\text{first orange and then blue})$:

$P(\text{orange on first spin})=\frac{1}{4}$ and $P(\text{blue on second spin})=\frac{1}{4}$, so $P(\text{first orange and then blue})=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\neq\frac{1}{8}$.

For $P(\text{first orange and then green})$:

$P(\text{orange on first spin})=\frac{1}{4}$ and $P(\text{green on second spin})=\frac{1}{4}$, so $P(\text{first orange and then green})=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\neq\frac{1}{4}$.

Answer:

$P(\text{both green})=\frac{1}{16}$