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Probability

by anb » Thu Sep 25, 2008 10:52 am
Hello,

Consider the following two questions:

(a)
In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble?

(b)
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

In question (a), you calculate the probability of picking a white marble first and then the blue marble and add it to the probability of picking a blue marble first, then picking the white marble. So you add two probabilities.

In question (b), you calculate the probability of picking a junior sibling pair (60/1000) and multiply by the probability of picking a senior sibling pair (1/800). Why don't you add the probability of picking the senior first, then the junior, as such in question (a)??? What's the fundamental difference between the two questions?

Thanks in advance!
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by Stuart@KaplanGMAT » Thu Sep 25, 2008 11:36 am
There are two fundamentally different concepts in probability:

1) Alternative probability; and
2) Multiple probability.

Your questions illustrate the difference between the two ideas.

Alternative probability questions ask us to calculate the probability of one of a series of different events occuring. Common words to help us identify alternative probability questions are "or", "at least" and "at most". Alternative probability also plays a role in multiple scenario questions (such as your question (a)), which I'll discuss below.

Here's your probability mantra #1:

Alternative = Add

Here's an example:

Bob and Fred both enter the same lottery. Bob buys 6 tickets and Fred buys 8 tickets. If there are a total of 100 tickets sold and only 1 prize is awarded, what's the probability that Bob or Fred wins the prize?

Prob(Bob) = 6/100
Prob(Fred) = 8/100

Prob(Bob or Fred) = 6/100 + 8/100 = 14/100

Multiple probability questions ask us to calculate the probability of more than one of a series of different events occuring. Common words to help us identify multiple probability questions are "and", "both" and "all". Multiple probability also plays a role in multiple scenario questions (such as your question (a)), which I'll discuss below.

Here's your probability mantra #2:

Multiple = Multiply

Here's an example:

A drawer contains 10 socks, 3 of which are red, 5 of which are blue and 2 of which are yellow. If Dawn pulls out 2 socks at random, without replacement, what's the probability that both are red?

Prob(1st red): 3/10
Prob(2nd red): 2/9
Prob(both red): 3/10 * 2/9 = 6/90

Multiple Scenario questions involve both alternative and multiple probability.

To calculate the probability of each individual option occuring, we multiply.

To calculate the probability of one of those options occuring, we add.

Here's an example:

Julia enters an essay contest in August and November. If she has a 10% chance of winning in August and a 20% chance of winning in November, what's the probability that she wins the contest in exactly one of the two months?

Well, there are 4 possible scenarios for the two months:

1) Lose both
2) Win August, Lose November
3) Lose August, Win November
4) Win August, Win November

The question requires that she win in EXACTLY one of the two months - scenarios (2) and (3) both qualify. Let's start by calculating the probability of each scenario.

Prob(Win August, Lose November) = (.1)(.8) = .08
Prob(Lose August, Win November) = (.9)(.2) = .18

We're happy with EITHER scenario, so we need to ADD the alternatives:

Prob(Win in exactly 1 month) = .08 + .18 = .26
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by stop@800 » Thu Sep 25, 2008 11:45 am
in first you
multiply the probability of red and the probability of white

so before addition you are actually doing the multiplication.
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by anb » Thu Sep 25, 2008 11:53 am
Stuart thanks for the elaboration! Very good explanation.


So, for question (b), why is the answer (60/1000) * (1/800), and not (60/1000)*(1/800) + (60/800)*(1/1000)? Question a considers both scenarios - picks white first, then blue; picks blue first, then white. Why not consider this for question (b)?
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by Stuart@KaplanGMAT » Thu Sep 25, 2008 11:57 am
anb wrote:Stuart thanks for the elaboration! Very good explanation.


So, for question (b), why is the answer (60/1000) * (1/800), and not (60/1000)*(1/800) + (60/800)*(1/1000)? Question a considers both scenarios - picks white first, then blue; picks blue first, then white. Why not consider this for question (b)?
There's only one scenario in question (b) - both students chosen are siblings.

If you add as you've done, you're double counting the same scenario. You could use either of your calculations to get the answer, but adding them up is counting the same scenario twice.
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by anb » Thu Sep 25, 2008 12:28 pm
thanks again!

so basically, for question (a) they are two different - a white ball and a blue ball. but for questions (b), sibling and sibling....
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by anb » Thu Sep 25, 2008 1:30 pm
sorry to add more, but I just want to make sure I understand...

so if your sock/drawer example were to ask, "What is the probability of choosing a red sock and a blue sock?"

Then the answer would be - (3/10)*(5/9) + (5/10)*(3/9)

Correct?
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by relaxin99 » Thu Sep 25, 2008 10:34 pm
So is the answer for a) the marbles question, 2/9?
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by anb » Fri Sep 26, 2008 11:33 am
yup, 2/9
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