How many times will the point at C and D in one play

[karush]

https://www.physicsforums.com/attachments/139
a game is played by rotating 2 arrows from the center of two circles per play with the pointer landing in one of the designated area's A B and C in one and E F and D in the other.
If played 120 times about how many times would the arrows land on the area of C and D in one play
wasn't sure of probablilty forumula to use for this but since it takes 3 tries to land in E and another 3 tries to Land in D
So 3x3=9 and 120/9 = 13.3 or about 13 times
so if this is correct. basically how is this set up with a probability is this a combination?
Thanks ahead(Cool)

 

[CaptainBlack]

https://www.physicsforums.com/attachments/139
a game is played by rotating 2 arrows from the center of two circles per play with the pointer landing in one of the designated area's A B and C in one and E F and D in the other.
If played 120 times about how many times would the arrows land on the area of C and D in one play
wasn't sure of probablilty forumula to use for this but since it takes 3 tries to land in E and another 3 tries to Land in D
So 3x3=9 and 120/9 = 13.3 or about 13 times
so if this is correct. basically how is this set up with a probability is this a combination?
Thanks ahead(Cool)

Assuming the arrows are spun so they land in a random area, the probability that on any spin the first arrow lands in C is 1/2 (as it accounts for 1/2 of all the directions the arrow could end up in)

The same for the second arrow and D.

As the results of the spins of the two arrows are independent the probability that arrow 1 ends in C and arrow 2 ends in D is the product of the individual probabilities and so is (1/2)(1/2)=1/4

So in 120 spins you would expect about 1/4 of then to have the arrows in C and D which is about 30 times.

CB

 

[HallsofIvy]

https://www.physicsforums.com/attachments/139
a game is played by rotating 2 arrows from the center of two circles per play with the pointer landing in one of the designated area's A B and C in one and E F and D in the other.
If played 120 times about how many times would the arrows land on the area of C and D in one play
wasn't sure of probablilty forumula to use for this but since it takes 3 tries to land in E and another 3 tries to Land in D
So 3x3=9 and 120/9 = 13.3 or about 13 times
so if this is correct. basically how is this set up with a probability is this a combination?
Thanks ahead(Cool)

IF all three areas in each spinner had the same area, THEN the probabilty of landing in C and D would be 1/3 so the probabilty of both landing in those areas would be 1/9 and your answer would be correct.
However, your picture shows that C and D each takes up 1/2 of the total area of the spinner, not 1/3.

 

[karush]

I see what you mean by possible 3rds however the problem gave this diagram and just asked probability of landing in C and D this is a realitively easy problem but I was wondering if this is probablitly combination (I don't know probability stuff hardly at all so trying to learn it)
looks like MHB is a good place to spend time... appreciatate much

 

[CellCoree]

The probability of landing on C and D in one play would depend on the size of the designated areas A, B, C, D, E, and F. Without knowing the exact sizes, it is difficult to determine the probability. However, assuming that all designated areas are equal in size, the probability of landing on C and D in one play would be 1/6, since there are 6 possible outcomes (A, B, C, D, E, F) and only 2 of them result in landing on C and D.

Therefore, in 120 plays, we can expect approximately 20 times (120/6 = 20) for the arrows to land on C and D. This is just an estimate and the actual number may vary based on the size of the designated areas and the accuracy of the rotation of the arrows.

As for the setup, this is a simple probability problem where we are trying to determine the likelihood of landing on a specific outcome (C and D) out of all possible outcomes. It is not a combination as there is no specific order or arrangement involved in this scenario. It is simply a matter of probability.