Probability Question

rbansal wrote:GMAT A bag holds 2 red marbles and 3 green marbles. If you removed two randomly selected marbles from the bag, what is the probability that both would be red?

ANSWER CHOICES

A 1/10
B 1/5
C 3/10
D 2/5
E 1/2


I eliminated D & E right away but the answer is A and I dont understand how they got there. Can you give me a detailed step by step explanation.

Hi,

when you select multiple items simultaneously, you can treat the question as though you select them one at a time, without replacement.

So, our question becomes:

If there are 2 red marbles and 3 green marbles in a bag, and you select a marble, then select a second marble without replacing the first, what's the probability that both marbles selected were red?

Now it's just simple probability:

prob = (# of desired outcomes)/(total # of possibilities)

Prob 1st marble red = 2/5
Prob 2nd marble red = 1/4

(without replacement, so we adjust the numbers as we go).

Since we want the probability that BOTH are red, we MULTIPLY each independent probability:

(2/5)(1/4) = 2/20 = 1/10

Presto!

The answer has to be 2C2/5C2.

1 / 10.

A.

Detailed explanation

No. of possible outcomes : From 5 marbles (2 red and 3 green), you can choose 2. That is 5C2 which is equal to 10.

No of desired outcomes: Both have to be red. From 2, select 2. There is only one way of doing this. That is to select both. (2C2)

So Probablity = Desired outcomes / All outcomes.

1 / 10.

Thank you so much this makes much more sense.

RB

@ Stuart,

On probability questions, I realized that you sometimes use the "one-minus" method and sometimes you don't. For example, on this question, you did not. I'm guess it was because it's not as complex as some of the other questions? So, do you have a recommendation on what type of probability question to use the "one-minus" method and when not to? I would appreciate your response.

jacob.t.jennings wrote:@ Stuart,

On probability questions, I realized that you sometimes use the "one-minus" method and sometimes you don't. For example, on this question, you did not. I'm guess it was because it's not as complex as some of the other questions? So, do you have a recommendation on what type of probability question to use the "one-minus" method and when not to? I would appreciate your response.

Great question!

The "one-minus" method is most useful on multiple scenario questions in which there are more scenarios that we want than we don't want.

The question will often have "at least" or "at most" in it, indicating that there are multiple possibilities that you want and don't want.

Thanks a lot Stuart!

Stuart, would you be kind enough to illustrate a probability problem with multiple scenarios. Just to get an idea of the one-minus method?

Vishnu88 wrote:Stuart, would you be kind enough to illustrate a probability problem with multiple scenarios. Just to get an idea of the one-minus method?

Sure! Let's use a very common GMAT probability question type: coin flips.

A fair coin is flipped 5 times. What's the probability that at least 1 of the 5 results is heads?

Whenever you see "at least" or "at most" in a question, you're dealing with multiple scenarios. Here, for example, we want "at least 1 head" out of 5 flips, so acceptable results are:

1 head and 4 tails;
2 heads and 3 tails;
3 heads and 2 tails;
4 heads and 1 tail; and
5 heads and 0 tails.

Since we want 1H4T OR 2H3T OR 3H2T OR 4H1T OR 5H01, one way to solve would be to figure out the probability of each scenario and ADD them together.

(Here's a very important rule to remember for probability and counting questions: to calculate MULTIPLE scenarios, MULTIPLY the individual ones; to calculate ALTERNATIVE scenarios, ADD the individual ones.)

However, if we're GMAT experts we notice that the only scenario that we DO NOT want is 0 heads and 5 tails. Since the sum of all probabilities is 1, we can use the "one minus" approach and set up the following equation:

Probability(what we want) = 1 - Probability(what we don't want)

Prob(at least 1 head) = 1 - Prob(0 heads)

0 heads is the same as 5 tails, and the probability of 5 tails is simply:

(1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

So:

Prob(at least 1 head) = 1 - 1/32 = 31/32

Stuart, thank you so much for that.

Couldn't have asked for anything better.

Cheers!