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GMAT PREP NOW - Probability Lesson 13

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GMAT PREP NOW - Probability Lesson 13

by [email protected] » Tue Mar 19, 2013 8:36 am
This lesson asks you two find the probability that the product of two selected numbers from a give set are positive. However, it seems to me that the way it is explained in the video determines whether the sum of the two numbers are positive, not the product. Has anyone else encountered this or am I just crazy?


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by Brent@GMATPrepNow » Tue Mar 19, 2013 9:06 am
This question requires us to use the probability rule.

In the solution we must find P(both numbers are positive OR both numbers are negative)
This equals P(both numbers are positive) + P(both numbers are negative)

So, we still need to find the probability that the product is positive, but since this requires to find P(event A OR event B), we need to find the sum of the two probabilities to do so.

This is covered in the GMAT Prep Now video #7 - Applying the OR Rule

I hope that helps.

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by J N » Tue Mar 19, 2013 2:41 pm
9/21?
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by Brent@GMATPrepNow » Tue Mar 19, 2013 3:05 pm
J N wrote:9/21?
Exacly

Here's the full solution.

P(positive product) = P(both numbers are positive OR both numbers are negative)
P(both numbers are positive) + P(both numbers are negative)

Now we'll calculate each part separately.

P(both numbers are positive)
P(both numbers are positive) = P(1st number is positive AND 2nd number is positive)
= P(1st number is positive) x P(2nd number is positive)
= (3/7) x (2/6)
= 1/7

P(both numbers are negative)
P(both numbers are negative) = P(1st number is negative AND 2nd number is negative)
= P(1st number is negative) x P(2nd number is negative)
= (4/7) x (3/6)
= 2/7

So, P(positive product) = 1/7 + 2/7
= 3/7

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by J N » Tue Mar 19, 2013 7:34 pm
Luck or another way???

overall possible
7!/2!5! = 21

Pick 2 negatives
4!/2!2! = 6

Pick 2 positives
3!/2! = 3

9 possible ways to get positive/21 possible combinations
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by Brent@GMATPrepNow » Tue Mar 19, 2013 7:42 pm
J N wrote:Luck or another way???

overall possible
7!/2!5! = 21

Pick 2 negatives
4!/2!2! = 6

Pick 2 positives
3!/2! = 3

9 possible ways to get positive/21 possible combinations
Great solution - perfect execution of counting techniques.

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by J N » Wed Mar 20, 2013 8:49 am
Is there a general guideline for when to use your approach vs combinations? I often get confused as to which method is preferred/better/easier/faster.......either method I never feel very confident in the answer.
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by Brent@GMATPrepNow » Wed Mar 20, 2013 8:52 am
That's a good question.

I'd say that using probability rules is typically the faster approach, but there's also a matter of comfort level. If your counting techniques are strong (which they appear to be, judging from your post in this thread), then stick with that.

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