Probability: Permutations or Combinations

[shikhargmat]

Although I do not find it difficult while solving the Probability questions, but do have a conceptual doubt.


While solving a Dice problem e.g. 2 dice are there. 1 showing Prime and other showing composite. Find probability.
In these problems we take 2 cases.
1) Prime on Die one and Composite on Die 2.
2) Composite on Die One and Composite on Die 2.
And add the results.
1/3 * 1/2 + 1/2 * 1/3
So in a way we can say that we are taking all the arrangements or combinations.


Whereas for Balls pickup problems such as this:
Find probability of picking 5 balls out of 8white and 4 Red (Condition: we must pick 3 white and 2 Red).
For such questions we solve as:
8C3 * 4C2 / 12C5
i.e. we took the combinations (and not permutations).

While solving, i know how to solve these kind of problems but still I find difficulty at conceptual level. Whether we ned to take all permutations or combinations?

[Stuart@KaplanGMAT]

Although I do not find it difficult while solving the Probability questions, but do have a conceptual doubt.


While solving a Dice problem e.g. 2 dice are there. 1 showing Prime and other showing composite. Find probability.
In these problems we take 2 cases.
1) Prime on Die one and Composite on Die 2.
2) Composite on Die One and Composite on Die 2.
And add the results.
1/3 * 1/2 + 1/2 * 1/3
So in a way we can say that we are taking all the arrangements or combinations.


Whereas for Balls pickup problems such as this:
Find probability of picking 5 balls out of 8white and 4 Red (Condition: we must pick 3 white and 2 Red).
For such questions we solve as:
8C3 * 4C2 / 12C5
i.e. we took the combinations (and not permutations).

While solving, i know how to solve these kind of problems but still I find difficulty at conceptual level. Whether we ned to take all permutations or combinations?

Hi,

precise wording is key on GMAT problems - I'm sorry to say that because of the grammar you've used, it's almost impossible to interpret the questions you're asking.

"2 dice are there. 1 showing Prime and other showing composite. Find probability."

just doesn't make any sense (there isn't even a question - find the probability of what?) - please pose your questions in full sentences.

[shikhargmat]

Q-1In a throw of two dice, find the probability of getting one prime and one composite number.

Q-2 A bag contains 8 white and 4 red bals. Five balls are drawn at random. hat is the probability that two of them are red and 3 are white?

I hope this is enough information.

[Stuart@KaplanGMAT]

Q-1In a throw of two dice, find the probability of getting one prime and one composite number.

Q-2 A bag contains 8 white and 4 red bals. Five balls are drawn at random. hat is the probability that two of them are red and 3 are white?

I hope this is enough information.

OK.. still some typos (take the time to check your posts!), but we can work with those questions.

Q1: the primes from 1-6 (assuming 6-sided dice) are 2, 3 and 5; the composites are 4 and 6. (1 is neither prime nor composite - 1 really is the loneliest number!)

As you noted, there are 2 cases:

prime/composite

and

composite/prime

Since we're happy with EITHER case, we ADD the probabilities together.

case 1: 3/6 * 2/6 = 6/36 = 1/6
case 2: 2/6 * 3/6 = 6/36 = 1/6

added together: 2/6 = 1/3

Q2:

also as you noted, we solve using the probability formula:

Probability = (# of desired results) / (total # of possibilities)

We have 8 white balls and want 3 of them, so there are 8C3 ways to choose 3 white balls; we have 4 red balls and want 2 of them, so there are 4C2 ways to choose 2 red balls; we have 12 balls total and want 5 of them, so there are 12C5 ways to choose 5 balls.

We're choosing 3 white AND 2 red, so we MULTIPLY to find our desired results:

Prob = 8C5*4C2/12C5

One key difference between these two problems is "replacement". For die roll (and coin toss) questions, we have an unlimited number of each outcome (just because you roll a 3 on the first die doesn't mean you can't roll a 3 on the next 17 dice as well); for the ball choosing question, once you select a particular ball you can't reselect that same one.

The second key difference is that in the second problem we aren't solving for alternative cases.

However, we could have solved either problem using either method.

For example, for Q1, we could have used combinations:

Prob = (# desired)/(total #)

Scenario 1 (prime then composite): 3C1/6C1 * 2C1/6C1 = 3/6 * 2/6 = 6/36 = 1/6
Scenario 2 (composite then prime): 2C1/6C1 * 3C1/6C1 = 2/6 * 3/6 = 6/36 = 1/6

and again, adding them together:

1/6 + 1/6 = 2/6 = 1/3

For Q2 it becomes far more time consuming and complicated to use probability, because all we end up doing is deriving the combinations formula - and if we know the formula, we might as well just plug in instead of creating the formula ourselves.