q 25.) a spinner has 4 equal sections; red, green, blue and orange. ricardo is going to spin the spinner…

q 25.) a spinner has 4 equal sections; red, green, blue and orange. ricardo is going to spin the spinner twice. he uses the chart below to help him determine the probability of an event. what is the probability that the spinner will land on a green section at least once? express your answer as a fraction in simplest form.\n1st spin 2nd spin outcomes\nred red red, red\nred green red, green\nred blue red, blue\nred orange red, orange\ngreen red green, red\ngreen green green, green\ngreen blue green, blue\ngreen orange green, orange\nblue red blue, red\nblue green blue, green\nblue blue blue, blue\nblue orange blue, orange\norange red orange, red\norange green orange, green\norange blue orange, blue\norange orange orange, orange\nanswer________

q 25.) a spinner has 4 equal sections; red, green, blue and orange. ricardo is going to spin the spinner twice. he uses the chart below to help him determine the probability of an event. what is the probability that the spinner will land on a green section at least once? express your answer as a fraction in simplest form.\n1st spin 2nd spin outcomes\nred red red, red\nred green red, green\nred blue red, blue\nred orange red, orange\ngreen red green, red\ngreen green green, green\ngreen blue green, blue\ngreen orange green, orange\nblue red blue, red\nblue green blue, green\nblue blue blue, blue\nblue orange blue, orange\norange red orange, red\norange green orange, green\norange blue orange, blue\norange orange orange, orange\nanswer________

Answer

Explanation:

Step1: Find the probability of not landing on green in one spin

The probability of not landing on green in one spin is $\frac{3}{4}$ since there are 3 non - green sections out of 4 total sections.

Step2: Find the probability of not landing on green in two spins

Since the spins are independent events, the probability of not landing on green in two spins is $\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$.

Step3: Find the probability of landing on green at least once

The probability of landing on green at least once is the complement of the probability of not landing on green in two spins. So it is $1-\frac{9}{16}=\frac{16 - 9}{16}=\frac{7}{16}$.

Answer:

$\frac{7}{16}$