My Math Forum Probability Question!! Help!

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 June 12th, 2011, 07:34 PM #1 Newbie   Joined: Jun 2011 Posts: 9 Thanks: 0 Probability Question!! Help! Let S = {a, b, c} be a sample space for a particular experiment. If outcome c is twice as likely to occur as outcome b, which is 3 times as likely to occur as outcome a, what is the probability that if you perform this experiment you will observe outcome c. How are you suppose to do this??
 June 12th, 2011, 08:12 PM #2 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Probability Question!! Help! A little algebra and substitution. First, always remember that all probabilities have to add up to 1.0. Therefore $a + b + c= 1.0$ Now let's look at the relationships. Since c is twice as likely to happen as b, then $c\=\ 2b$. And since b is three times as likely to happen as a, then $b\=\ 3a$. Now start substituting into the original equation: $a\ +\ [3a]\ +\ c\=\ 1$ That substitutes in my equivalent for b: three times a. But now what do we do with c? Well, we really don't want to substitute in $c\=\ 2b$ because we'd still have a b in there and have two variables. But we can substitute in the equivalent of b (in terms of a) that we have. That is $c\=\ 2[3a]$. This puts my "3a" in where the b was. So this is what we can now use for the c: $a\ +\ 3a\ +\ 2[3a]\=\ 1.0$ It's a series of substitutions. And if you can solve that from there, you'll have a. You can then use the relationships to get b and c. And then check them to make sure they add up to 1.
 June 12th, 2011, 08:36 PM #3 Newbie   Joined: Jun 2011 Posts: 9 Thanks: 0 Re: Probability Question!! Help! Thank you soo much! I actually did that but I made some arithmetic error...... 1 -6a - 3a isn't 1 - 3a. It's 1 -9a. Learnt that the hard way.
 June 12th, 2011, 08:50 PM #4 Newbie   Joined: Jun 2011 Posts: 9 Thanks: 0 Re: Probability Question!! Help! Hey I'm also stuck on the next part to this question. I've been working on it for so long but I have no idea how to solve it. Can someone please help? Thanks! Let S = {a, b, c} be a sample space for a particular experiment. Furthermore, let E be the event that outcome a is observed, and F be the event that outcome c is not observed. If event F is twice as likely to occur as event E, and event E is 4 times as likely to not occur as it is to occur, how much more likely is outcome c to occur than outcome b?
June 13th, 2011, 05:38 AM   #5
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Re: Probability Question!! Help!

P(X) is defined as "the probability that event X will happen. Try rewrite your excersize in such statements.

Quote:
 Originally Posted by Azntopia Furthermore, let E be the event that outcome a is observed,...
P(E)=P(a)
Quote:
 Originally Posted by You ...and F be the event that outcome c is not observed....
P(F)=P(not c)=P(a or b)
Quote:
 Originally Posted by You ...If event F is twice as likely to occur as event E,...
P(F)=2*P(E)
Quote:
 Originally Posted by You ...and event E is 4 times as likely to not occur as it is to occur,...
P(not E)+P(E)=1 (always)
4*P(E)=P(not E)

Now, try to figure out what you need to find
Quote:
 Originally Posted by You ...how much more likely is outcome c to occur than outcome b?

 June 13th, 2011, 06:02 AM #6 Newbie   Joined: Jun 2011 Posts: 9 Thanks: 0 Re: Probability Question!! Help! Hey Hoempa thanks for the help, but i'm still having some trouble trying to solve this question. I still don't know how to really start this problem even given the statements.
 June 13th, 2011, 06:27 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Probability Question!! Help! Consider P(b) = 2P(c) = 3P(a) and P(c) + P(b) + P(a) = 1. You can make the system 3P(c) + P(a) = 1, 2P(c) - 3P(a) = 0 ? 11P(c) = 3 ? P(c) = 3/11.
 June 13th, 2011, 01:10 PM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Probability Question!! Help! On top of that: do you understand how to find such statements with the given information? Assuming you understand the question [color=#0000FF]Erimess[/color] helped you on, I'll give you another start. P(not E)+P(E)=1 (always) 4*P(E)=P(not E) These help you to find P(E) by making a substitution. P(F)=2*P(E) You can substitute some more in this. P(F)=P(a or b) and P(E)=P(a). What can you find? Now, you can find P(a) and P(b). How to find P(c)? Now, check with all the statements. Are they all correct with te answers?
 June 14th, 2011, 01:48 PM #9 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Probability Question!! Help! OK... I tend to look at things as pictures. I can put it into proper equations and math symbols, but I like to literally see it in my mind first to see a relationship between things. Let's start with E, cause personally I think that's where you need to start and from there you should be able to get some other things. Here is what I "see" with E: it is four times as likely NOT to happen as to happen, and here's the picture I see in my mind, literally: Sorry, a little crude - I just did that in Paint real quick. So the "not occur" is 4 parts, and the "occur" is one part, cause the "not" is 4 times as likely, so there has to be 4 of "not occur" to 1 of "occur." Now looking at this, and remembering that "not" means something besides E happens but it still has to all add up to 1, what portion is the "occur"? Once you have that, you can get F. Now, if you can get the P(E), then you'll have P(a). Remember that "not c" means either a OR b, so if you get P(E), you can get P(F), which is also P(a or b). Since you have P(a) and P(a or b), what then is P(b)? Then go from there with the rest of the info. And never forget they all add up to 1 - that's imperative. Give it a try. By the way, you stated in your other post to me that you had done it that way but basically just made a math error. If you always show us what you tried to do, then we can see what you're getting, what you're not, and hone in to where the mistakes are to clear those up. So maybe try posting what you have attempted on this and see where things are at.
June 14th, 2011, 01:51 PM   #10
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Re: Probability Question!! Help!

Quote:
 Originally Posted by Erimess OK... I tend to look at things as pictures...
A right brainer!

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