Probability of Simultaneous Events - Veritas vs MGMAT

Question

Hi,

Background: I am studying probability, one of my weakest areas, using both MGMAT and Veritas books.

The probability of a simultaneous events (picking two balls without replacement) is explained DIFFERENTLY in those books. I am turning to wise Clubbers to point out which is right.

Question: there are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles? (MGMAT Word Translation, p93, Q9)

The correct answer, according to MGMAT is 2/9. You can pick first a white marble and then a blue marble which is 1/5 x 5/9 = 1/9. You can also pick the blue first and then white: 1/2 x 2/9 = 1/9.

MGMAT then urges to add up both probabilities and get 2/9 as the correct answer.

In the Veritas Combinatorics, the correct answer is 1/9 i.e. in simultaneous events, you do not add up the both cases.

My hunch is that Veritas is correct. I think this, because the probability of a simultaneous event (or event without replacement) is ALWAYS THE SAME regardless of the order (i.e. if you pick white or blue marble first). I think that if you add up, you are overestimating the probability of the event.

As said, I am weak at probability, and want to hear what other, wiser Clubbers say! Please let me know which book got it right...!

roshpat · Answer

Thanks very much, Karishma for the clarification and the example. I missed the "simultaneous".
I now understand the difference between "simultaneous" and one after the other. These are both for picking without replacement.

Basically, when they are picked one after the other - the order matters and hence you need calculate the number of ways this can be done and multiply with the probability.

In the simultaneous case, I had one quick question- if the order does not matter, why does this not reduce to a combinations problem? (As in the case of your example and MGMAT, this does not happen as that method leads us to the 1st case, one after the other).

Is this because of the conditional dependence?

Sorry to raise so many questions but want to make sure I at least get the hang of these questions.

Thanks very much.

eragotte · Answer

With this method why do supplementary probabilities not need to add up? I would never say the probability of grabbing two aces similtaneously out of a deck of cards is 1/221 but the probability of not grabbing two aces is 150/221... That does not make sense to me. Anyway you can link me to where mathematicians prove this, with an explanation as to why supplementary probabilities can not add up to 1?

Bunuel · Answer

MGMAT answer is correct.
 There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

P(WB)=2*2/10*5/9=2/9, we multiply by 2 as the scenario WB can occur in two ways: first ball is white and the second is blue OR first ball is blue and the second is white.

I don't think the above is correct.

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

So the answer for both questions below is 1/3.

Case 1: There is one red, one blue and one green ball. I pick two balls. What is the probability that one is red and one blue?

P=2*1/3*1/2=1/3.

Case 2: There is one red, one blue and one green ball. I pick two balls  simultaneously . What is the probability that one is red and one blue?

You can do the same as above or consider this: P=# of favorable outcomes/total # of outcomes --> there are total of 3 outcomes possible for 2 balls: (Red, Blue), (Red, Green), and (Blue, Green) --> # of favorable outcomes is 1: (Red, Blue) --> P=1/3 (or  P=\frac{C^2_2}{C^2_3}=\frac{1}{3} ).

roshpat · Answer

Thanks very much, Bunel.

Would I be correct in assuming the following:

1.Picking the gumballs simultaneously can be considered as a combination problem as the order does not matter.
2.One after the other can be considered as a permutation problem as the order matters.

Either way, the answer will be the same.

Bunuel · Answer

I'm not sure understood what you mean but you can solve below problem using 2 approaches.

A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

Direct combinatorial approach: 
 Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes} :

Favorable outcomes:  C^1_7*C^1_5*C^1_4 ;
Total # of outcomes:  C^3_{14} ,;

P=\frac{C^1_7*C^1_5*C^1_4}{C^3_{14}}=\frac{1}{4} .

Direct probability approach: 
 P=3!*\frac{7}{16}*\frac{5}{15}*\frac{4}{14}=\frac{1}{4} , we multiply by 3! as the scenario BGR can occur in several ways: BGR, BRG, GBR, GRB, RBG, RGB (basically # of permutations of 3 letters BGR which is 3!=6).

Again in doesn't matter whether the machine dispenses these gumballs simultaneously or one at a time.

Hope it helps.

KarishmaB · Answer

Picking Red and Blue simultaneously is the same as picking first Red and then Blue without replacement. It is not the same as picking any one and then the other without replacement (where we consider Red Blue and Blue Red)

So the answer in both the cases will not be 1/3 (as I explained before)

These are dependent events.
P(R and B) = P(R)*P(B given that R has occurred) = 1/3 * 1/2 = 1/6

For more on the same, please check out Probability of Simultaneous Events. You can use Veritas Prep Combinatorics and Probability book or any other standard book.

Bunuel · Answer

1/6 would be the result if we were asked to find the probability that FIRST ball is red and SECOND is blue. But the question asks "what is the probability that one is red and one blue?" and in this case no matter whether they are picked one at a time (without replacement) or simultaneously the answer is the same 1/3.

Anyway: are you saying that there is a difference whether we pick two balls simultaneously, or pick them one at a time (without replacement)?

KarishmaB · Answer

No. There isn't. Probability of picking a red and a blue ball simultaneously is the same as picking a red and then a blue ball without replacement (or a blue and then a red ball without replacement). But you do not have to consider both cases - red and then blue, blue and then red. Only one of them and any one of them. 
They are dependent events and this is conditional probability. As I said before, one can read about this in Veritas book (and perhaps MGMAT too as suggested by someone in a previous post)

Bunuel · Answer

Karishma there must be some misunderstanding.

Two questions:

Case 1: There is one red, one blue and one green ball. I pick two balls. What is the probability that one is red and one blue?

Case 2: There is one red, one blue and one green ball. I pick two balls simultaneously. What is the probability that one is red and one blue?

I'm saying that these are two identical questions and the answer is 1/3 for both, because there is no difference whether we pick two balls simultaneously, or pick them one at a time (without replacement). Also in neither of questions we have a case of conditional probability.

Next, if you still say that the answer to the second question:  "There is one red, one blue and one green ball. I pick two balls simultaneously. What is the probability that one is red and one blue?"  is 1/6 then can you please give the breakdown of the the rest 5/6?

As for me when you pick two balls out of three you can have in your hand:
1. Red and Blue;
2. Red and Green;
3. Blue and Green.

Only 3 cases and each is equally likely, the probability of one is red and one blue (scenario #1) is 1/3.

Again: 1/6 would be the result if we were asked to find the probability that FIRST ball is red and SECOND is blue.

eragotte · Answer

This is what I was saying. I was going mental last night trying to figure out how the heck they could be different, supplementary probabilities must add to 1.

KarishmaB · Answer

Probabilities need to add to 1 only when the possible cases are mutually exclusive and exhaustive. So do not worry about it in conditional probability scenario.

Simultaneous events represent joint probability. It is given as
P(A and B) = P(A given B)*P(B)

P(A given B) is the conditional probability that A occurs given B has occurred.

GMAT questions related to simultaneous events can be worked out the way I have suggested in my previous posts. That is all that is needed.
and I repeat, it is a tricky concept which is not intuitive but has been proved by mathematicians.
and yet again, if you are in interested in more details, please check out a standard text book for probability of simultaneous events, conditional probability, joint probability, dependent events etc.

Bunuel · Answer

I think that we do have mutually exclusive and exhaustive events:

There are 3 mutually exclusive events when picking 2 balls out of 3 (one red, one blue and one green ball):
1. Red and Blue;
2. Red and Green;
3. Blue and Green.

And these outcomes are exhaustive events too, what other outcomes can you have?

Also we do not have conditional probability here. The question asks: the probability that one is red and one blue, not the probability of second blue when first red.

IanStewart · Answer

Bunuel asked me to chime in here. Wow, this thread has become confused!

This is all actually very simple. Say I have a bag with 3 red and 2 green marbles, and I stick both of my hands in the bag and grab hold of two marbles. Obviously the probability that I pick 1 red and 1 green marble can't possibly be any different if I take both my hands out at once, or if I take them out one at a time (edit - to take a real world example, if a poker player is dealt two cards, she isn't less likely to have a pair of aces if she looks at them one at a time than if she looks at them both simultaneously). The distinction some people are drawing in posts above between 'simultaneous selections' and 'selections one at a time' is mathematically irrelevant. You can take either perspective when you answer these kinds of questions, and you'll always get the same answer (provided you do the math correctly!).

If we go back to the question in the original post:

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

the answer here is certainly 2/9. You can either think of making the selections one at a time, in which case there are two distinct ways in which we get the required result:

(white AND blue) OR (blue AND white)

(2/10)(5/9) + (5/9)(2/10) = 2/9

I'd find this the easier way to look at the problem. Or if you prefer to think, in mathematical terms, of everything occurring at once, there are 10C2 = 45 ways of picking two marbles in total, and there are (2C1)*(5C1) = 10 ways of picking a blue marble and a red marble, so the answer must be 10/45 = 2/9. If a book is claiming the answer to this question is 1/9 (or if the book is suggesting you use some method that would lead to the answer 1/9), the book is simply wrong. If you were asked explicitly: "Bob will take one marble out of the bag, and then, without putting the first marble back, will take a second marble out of the bag. What is the probability the first marble is white and the second marble is blue?" then the answer is 1/9, of course. But that's a different question from the one above.

This is simply false. You can see that we can't simply take one particular order and use it to find the answer to a question like the above by using the simplest possible example. Say you have 1 red and 1 blue marble in a bag, and you pick two marbles without replacement. What's the probability you get 1 red and 1 blue marble? Obviously it's 100%, and not 1/2; you can't choose the order 'first Red and then Blue' and hope to get the right answer.

Now, picking Red and Blue simultaneously *is* the same as picking first Red and then Blue ***OR*** picking first Blue and then Red. But you absolutely need to add the probabilities from the two cases or else your answer will be half what it should be.

Bunuel · Answer

Thank you very much Ian!

I was running out of words to explain this simple concept.

roshpat · Answer

Thank you very much, Ian and Bunel for explaining these very clearly.

mbafall2011 · Answer

This one was very similar to the other question where i had the exact same doubt but i solved this question right and got that wrong initially~! selecting-3-balls-36702.html

144144 · Answer

GEEZ... giants at work. thanks all.

almostfamous · Answer

Wow. Messy discussions, beautifully corrected!

I started worrying that I initiated a thread that became a messy monster...glad that maths gurus stepped up to untangle the mess!

Thank you!

a03b · Answer

Excellent excellent thread! First, I read ONLY page 1. I was going crazy thinking how the hell the answer could be 1/9. :D

Then I realized, there is a second page. ALERT EVERYBODY! The answer is 2/9!!! Read page 2.

manmohansingh3010 · Answer

Hi, here is a question, I tried to think of a solution, but it is not working out, kindly let me know why?

A box contains 5 green, 4 yellow and 3 white marbles. 3 marbles are drawn at random. What is the probability that they are not of the same color?

Sol(as thought out by me) : (5C1*4C1*3C1)/12C3

Also please let me know, this kind of solution will be applicable in which case? And how is drawing at random different from drawing one after the other?

Probability of Simultaneous Events - Veritas vs MGMAT

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