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Ludmilka [50]
2 years ago
12

The spinner below is spun twice. If the spinner lands on a border, that spin does not count and spin again. It is equally likely

that the spinner will land in each of the six sectors.
RED
RED
RED
BLUE
BLUE
CYAN
[Graphs generated by this script: initPicture(-100,100,-100,100);stroke='red'; fill='red';arc([90,0],[45,77.94228],90);path([[0,0],[90,0],[45,77.94228],[0,0]]);arc([-45,77.94228],[-90,0],90);path([[0,0],[-90,0],[-45,77.94228],[0,0]]);arc([-45,-77.94228],[45,-77.94228],90);path([[0,0],[-45,-77.94228],[45,-77.94228],[0,0]]);stroke='blue'; fill='blue';arc([45,77.94228],[-45,77.94228],90);path([[0,0],[45,77.94228],[-45,77.94228],[0,0]]);arc([-90,0],[-45,-77.94228],90);path([[0,0],[-90,0],[-45,-77.94228],[0,0]]);stroke='cyan'; fill='cyan';arc([45,-77.94228],[90,0],90);path([[0,0],[45,-77.94228],[90,0],[0,0]]);stroke='black'; fill='none';circle([0,0],90);line([45,77.94228],[-45,-77.94228]);line([-45,77.94228],[45,-77.94228]);line([-90,0],[90,0]);fontfamily='helvetica';fontfill='white';fontstyle='normal';fontweight='bold';text([55,25],'RED');text([-50,25],'RED');text([0,-70],'RED');text([0,65],'BLUE');text([-55,-30],'BLUE');fontfill='black';text([50,-25],'CYAN');strokewidth='5';line([-35,-35],[30,30]);line([25,15],[30,30]);line([15,25],[30,30]);]

For each question below, enter your response as a reduced fraction.

Find the probability of spinning blue on the first spin and cyan on the second spin.


Find the probability of spinning red on the first spin and cyan on the second spin.


Find the probability of NOT spinning red on either spin. (Not red on the first spin and not red on the second spin.)
Mathematics
1 answer:
cestrela7 [59]2 years ago
3 0

Answer:

A) 1/6 & 1/12 (with borders) 1/3 & 1/6 (without borders)

B) 1/4 & 1/12 (with borders) 1/2 & 1/6 (without borders)

C) 1/4 & 1/4  (with borders) 1/2 & 1/2 (without borders)

Step-by-step explanation:

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