Stuck on math problems ]:?

Question 12:

A circle has 360 degrees in it, so define the sections by that.

Section y is half, so 180 degrees there

X is listed as 120 degrees

That leavez z as 360-180-120=60 degrees

In short:

Z is 60 degrees of the circle, so the spinner has a 60/360 or 1/6 chance of hitting section z. or 16.666...% chance.

Y is 180 degrees of the circle, so the spinner has a 180/360 or 1/2 chance of hitting section Y. or 50%

X is 120 degrees of the circle, so the spinner has a 120/360 or 1/3 chance of hitting section X. or 33.333...% chance.

You can also put things in simple uniform terms if you want. The spinner has a 1/6 of hitting Z, 3/6 chance of hitting Y, or 2/6 chance of hitting X

For the subquestions asking what are the chances of not hitting an area, then it's the opposite. The spinner has a 5/6 chance of not hitting Z, 3/6 of not hitting Y, and 4/6 chance of not hitting X.

When you put things in uniform terms, it makes it easy to sort out.

a) Hitting X: 1 in 3 chance

b) Not hitting X: 2 in 3 chance

c) Getting Y or X: Y is 3/6, X is 2/6, so there a 5 in 6 chance.

d) Getting Y and Z at the same time: There's only one point in the circle where they connect. Technically, I think they are looking for a 1 / 360 chance, but truthfully, the chances are a LOT smaller then that, has the chance of getting the EXACT point and the length of the line that would allow this to occur would play a factor there.

e) Hitting z: 1 in 6 chance

f) No hitting Z: 5 in 6 chance

Now for question 13:

a) for the same reason that I mentioned of the spinner not hitting a section. The chance of something happening plus the chance of something not happening = 1. After all, either something happens or it doesn't. There's not alternate universe here. Let's use the spinner from the previous question as an example. The chance that the spinner hits Z is 1/6, the chance that the spinner does NOT hit Z is 5/6. Add the two possibilites 1/6 + 5/6 = 6/6 or 1. Therefore if you know the odds of an event, then 1 minus that fraction are the odds of that event not occurring.

b) A deck has 52 cards and four aces. So you have a 48/52 chance of not drawing an ace. That fraction reduces to 12/13 chance or a 92.307...% chance

c) This one you can't really put into a finite number as the percentage is based on the size of the thumbtack's head. But to calculate that you would need 1 - (percent of the head of the thumbtack's head as compared to the circumfrance of the circle.) I know that's a bit vague, but that is about as good as that is going to get with this question in my opinion.

d) You actually have four possibilities here. Each coin has to possibility Heads and tails, so 2 chances. And there are two coins. So there are four possibilites (2 coins * 2 states). So you have a one in four chance. Think of it this way. H=Head, T=Tails. So you possibliies are HH, HT, TH, TT.

e) Assuming we are talking about a six sided dice (there are other types of dice out there), then this is simple. 1 in 6 chance.

f) Assuming that the car key is the same approximate size and weight, then you have a 1 in 5 chance of not selecting the right key.