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 April 13th, 2012, 01:09 PM #1 Member   Joined: Nov 2011 Posts: 43 Thanks: 0 Probability Question Need help D; Q: 30% of people at a baseball game bought cotton candy and water. And half of the people who bought water also bought cotton candy. Determine the probability of people who bought cotton candy. do you just take 0.3 and divide it by 0.5? D; and the answer's 0.6? someone tell me I'm right
April 13th, 2012, 02:30 PM   #2
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Re: Probability Question

Hello, Chee!

Quote:
 30% of people at a baseball game bought cotton candy and water. And half of the people who bought water also bought cotton candy. Determine the probability of people who bought cotton candy. Do you just take 0.3 and divide it by 0.5? and the answer's 0.6? Someone tell me I'm right.[color=beige] . . [/color][color=blue] . . . sorry![/color]

There is insufficient information.

Did everyone at the game buy either cotton candy or water (or both)?
If so, we can solve the problem.

Consider this Venn diagram.

Code:
              * * *   * * *
* Candy * Water *
*       *   *       *
*       *30%*  30%  *
*       *   *       *
*       *       *
* * *   * * *
30% bought both Water and Candy.
Half of those who bought Water also bought Candy.
Hence, the other half is: those who bought Water only (30%).
Then 40% bought Candy only.

Therefore: 70% bought Candy.

April 13th, 2012, 02:58 PM   #3
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Re: Probability Question

Quote:
 Originally Posted by soroban Therefore: 70% bought Candy.
omg why.. -______-" can't you use the formula, $P( A and B )= P(A) * P (B I A) ?$ and sub in 0.3 for P(A AND B) and 0.5 for P(B I A) and isolate for P(A) ? -___-"?

April 16th, 2012, 11:45 PM   #4
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Re: Probability Question

Quote:
Originally Posted by Chee
Quote:
 Originally Posted by soroban Therefore: 70% bought Candy.
omg why.. -______-" can't you use the formula, $P( A and B )= P(A) * P (B I A) ?$ and sub in 0.3 for P(A AND B) and 0.5 for P(B I A) and isolate for P(A) ? -___-"?
Because you don't know the numbers. Let's make this easier by using W and C instead of A and B, cause I have no idea what your A's and B's stand for. If the probability of B given A is .5, then your A is the water and you already know that, so why solve for it? But just to make sure you understand, let's use the W and C and go through it:

$P(C|W)\=\ \frac{P(C\ \cap\ W)}{P(W)}$

$.5\=\ \frac{.3}{.6}$

Great, that statement's true. And it gets you nowhere cause you already know the probability of W anyway, so there isn't anything to solve for.

Now let's do it the way you seem to be attempting to do:
$P(W|C)\=\ \frac{P(W\ \cap\ C)}{P(C)}$

$\=\ \frac{.3}{?}$

The problem here is that you don't know W|C. It's not .5, because that's C|W. The problem is that you think .5 is W|C when it's not. You don't know that number.

It might help to understand what a conditional is, rather than trying to plug n chug numbers when you don't know which is which. By saying what's the probability of someone buying water given they have already bought candy, you are isolating down to only those people who bought candy. That is, you're isolating to the left circle in the Venn diagram. You toss out everyone else, so only the left circle exists now. So the probability of any of those people buying water is only the intersection people, out of the people who bought candy. To get the probability of the 30% in the intersection, out of the people in the left circle, you divide the intersection by the left circle. (Basic percentage rules, and where the equation comes from.) But you don't have a total for the left circle to divide by. .3 intersection divided by what? So you can't solve for that, in order to plug it in and then solve for the "what."

April 16th, 2012, 11:46 PM   #5
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Re: Probability Question

Quote:
 Originally Posted by Chee do you just take 0.3 and divide it by 0.5? D; and the answer's 0.6? someone tell me I'm right
I'm curious how you came up with this?

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