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Solve with combinations

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Solve with combinations

by Jinglander » Tue Sep 28, 2010 8:32 am
I know how to solve this using a probability decision tree, but wanted to see if it could be solved with combinations.

Renee, has a bag of 6 candies, 4 are sweet 2 are sour. Jack picks two candies simultaneously and at random, what is the chance that exactly one candy he picked is sour.

I know the total combinations its just 6!/2!4!. But how do I find the number of outcomes that have exactly one sour. Clearly the numbers are small enough that I can just count the number of out comes but i want a formula that can be applied to larger numbers
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by Geva@EconomistGMAT » Tue Sep 28, 2010 9:24 am
Caveat: the method below is a rather long and complicated approach, but that's what you asked for. What I would do with your original problem is find probability one step at a time. Split into separate events: first choice, second choice.
Probability of exactly one sour means two scenarios: sour only on first, or sour only on second.
P(sour on first choice) 2/6 * P(not sour on second choice) 4/5 = 8/30.
OR
P(not sour on first choice) 4/6 * P(sour on second choice) 2/5 = 8/30
Thus, final probability is 8/30+8/30 = 8/15.

XXXXX

Re: finding the number of wanted outcomes using combinations:
Deal with each group separately, then multiply. To get exactly one sour, you want 1 sour (out of 2), and one sweet (out of 4). thus, there are 2C1 =2 ways of choosing sours, and 4C1=4 ways of choosing sweet. Multiply the two groups to find the number of ways of choosing your wanted outcomes.

For example, the same question with 4 sweet and 3 sours, where we need to choose 4 candies, prob of 2 candies sour:

Total number of outcomes: choose 4 out of 7, or 7!/3!4!

number of wanted outcomes - you basically want two sour, two sweet. deal with each group separately and multiply:
number of ways of choosing 2 sour out of 3: 3!/1!2!=3
number of ways of choosing 2 sweet out of 4: 4!/2!2! = 6
Thus, total number of wanted outcomes is 3*6=18.
(Basically, for each of the three ways of choosing the sours, there are 6 ways of choosing the sweets - hence 3*8).

Caveat2: You're not likely to see a question this complicated on the real GMAT.
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by diebeatsthegmat » Tue Sep 28, 2010 9:25 am
Jinglander wrote:I know how to solve this using a probability decision tree, but wanted to see if it could be solved with combinations.

Renee, has a bag of 6 candies, 4 are sweet 2 are sour. Jack picks two candies simultaneously and at random, what is the chance that exactly one candy he picked is sour.

I know the total combinations its just 6!/2!4!. But how do I find the number of outcomes that have exactly one sour. Clearly the numbers are small enough that I can just count the number of out comes but i want a formula that can be applied to larger numbers
if you this way, you will have to find the probability to find 1 sweet and 1 sour to get exact 1 sour candy.
the way to select 1 candy is 4!/3!1!=4 and select 1 sour candy is 2!/1!1!=2
total ways to select 1 sweet and 1 sour candy is 2*4=8
thus the answer is 8/15 ( 15 is the combination you found from 6!/2!4!

you can solve the problem by another different way
select 1 sweet candy and 1 sour candy or select 1 sour first and 1 sweet candy later
thus the probability will be 4/6*2/5=8/30 and you must multiple 2 because there are 2 times depending which kind of candy you selected, sour or sweet
the answer is also 8/15
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