Probability of Simultaneous Events - Veritas vs MGMAT

No, you are right, that it is 1/9 overall.

Heres the explanation for 10:

Let's say I pick a white ball - then I have five choices for picking the second ball so its blue. Similarly for the second white ball - there's 5 choices. But overall, both the white balls are replaceable with each other, so the final answer is divided by 2 to give you 1/9.

Thanks!

I re-read MGMAT, and the book actually does explain the logics of simultaneous event probability correctly (under "The Domino Effect", Word Translations p89 - -90). It also shows a correct solution for the example it uses on p90.

The problem I posted appears in the Problem Set of the Probability Chapter, and Question 9 in it, as we discovered here, appears to have the wrong solution. With your help, now I am confident how to calculate the probability of a simultaneous event.

Thanks again for helping me to figure it out!

Let me add a note here:

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

Probability of picking 1 W and 1 B ball simultaneously = Probability of picking a W and then a B without replacement = Probability of picking a B and then a W without replacement

Probability of picking a W and then a B without replacement = 1/5 * 5/9 = 1/9

Probability of picking a B and then a W without replacement = 1/2 * 2/9 = 1/9

So whenever you need to pick two things simultaneously, give them your own order and calculate in any one way.

Karishma,

Thanks a lot for the detailed explanation.
After having studies Probability with both MGMAT and Veritas, I have to say that I find the latter much more comprehensive. In fact, the Veritas Combinatorics is the FIRST and ONLY material that made me feel comfortable with the topic - finally!

It may not be fair to directly compare these two materials (after all, MGMAT packs Probability, Combinatorics and a bunch of other advanced topics under 'Word Translations', wheres Veritas devotes an entire book for Combinatorics/Probability). But I am really grateful that I found Veritas!

Cheers,

almostfamous: I am glad the book helped. It is definitely comprehensive and covers everything that is relevant for GMAT.

Wanted to thank you all for raising and providing your insights to address these type of problems. I just had a quick question - Say if we were to extend this to more than 2 marbles being picked for example:

If the question were : There are 3 red, 2 white, and 5 blue marbles. If Bob takes 4 marbles out of the bag, what is the probability that he will have 2 red and 2 blue marbles?

In this case, the same concept can be extended

Probability of picking R, R, B, B (w/o replacement) in this order - 3/10 * 2/9* 5/8* 4/7

Probability of picking R, B, R, B (w/o replacement) in this order - 3/10 * 5/9 * 2/8 * 4/7

and the other different ways of picking these 4 marbles..

Again, order does not matter and we would only need to keep track of the following:

1. The total number of marbles which reduce with each pick
2. The remaining marbles in the color that has been picked earlier.

Would appreciate it if someone could tell me if my reasoning is correct.

Thanks very much.

Your logic is correct.

Regardless of the number of marbles to be selected, the underlying logic for this type of questions hinges on the same one as the one for simultaneous picking.

In your example:
Red: 3
White: 2
Black: 5
Total: 10

As we are picking the marbles "simultaneously" (i.e. without replacement), the probability is:

3/10 x 2/9 x 5/8 x 4/7 = 1/42

Hope this helps!

Thanks very much for the very quick reply. Glad to know that I am on the right track!! Perm, Comb & Probability were my weakest subjects at school!!

Intuitively I am missing something reading this thread. I can't understand something unless I understand why it is true.

Consider a simpler example to explain my confusion.

There are 3 coins in a bag. A quarter, a nickel and a dime. What is the probability that when reaching into the bag and pulling out 2 coins you get 30 cents.

By your method, 1/3 chance * 1/2 chance = 1/6 chance that you get a nickel and a quarter.

This is clearly wrong though as there are only 3 combos -> nickel dime, nickel quarter, dime quarter. Hence, by common sense the answer is 1/3.



*** similarly in the example given:

Blue white 1/2*2/9 = 1/9
Blue red 1/2*1/3 = 1/6
White red 3/10*2/9 = 1/15
Blue blue ½*4/9 = 2/9
Red red 3/10*2/9 = 1/15
White white 2/10*1/9 = 1/45
Sums 59/90
Where as I say the first three should be doubled because they can happen in 2 different ways, therefore 90/90 sum

The original problem deals with a problem category that could be representably called "probability with no replacement".

Your example changes the problem category; it is a combinatoric problem:

How could you pick 30 cents by picking two coins out of 1 N, 1 Q, and 1 D? This is the same as: what is the probability of picking both N and Q? As Picking N and then Q or Q and then N does not affect the total sum, the order does NOT matter.

In this, the total outcome is: 3!/(3-2)!*2* (because order does NOT matter) = 3
Favorable outcome: one out of 3 (N & Q, order does not matter)

Probability is 1/3.

Eh I am not convinced. For a probability problem all possible outcomes have to cover the full probability.

For this question the probability of not picking a white ball is clearly 8/10*7/9 = 0.622

both white can be given by 2/10*1/9 = 0.0222

White and blue 2*(1/2)*(2/9) = 0.222222 = 2/9

white and red 2*(1/5)*(3/9) = 0.13333

= 1.0


In the end call it what you like, but if you ran this experiment 1,000,000 times, very close to 222,222 trials would consist of 1 blue and 1 white.

I see that you are solving this problem by calculating the supplementary probability.

As the problem asks for the probability of an exact scenario (2 R and 2 W), instead of "at least one red" etc., it is easier to approach it by directly calculating the exact probability asked.

Sorry, I am going to raise one more quick question with you. I am just going over the MGMAT questions after going through Veritas and am a bit muddled.

There is an example in the MGMAT book -

"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?"

In this case, if we are to apply the Veritas logic, that the order does not matter, simply:

7/16* 6/15*4/14 = 1/24

MGMAT goes on to say that there are 3! ways to calculate this and multiplies 1/24 with 3! to get 1/4.

Just wanted to check - this is an overestimate right and the correct answer is 1/24, isnt it?

Would really appreciate a clarification on this point.

Thanks very much.

Would you post the exact page of the MGMAT book where I can find this problem? I own both the MGMAT Word Translations and Veritas Combinatorics. If I can see the problem on the book, it will help me to understand your question better.

Thanks.

I would write it on the test as 1/4 as I wrote in my last few posts. Try figuring out supplementary probabilities to see if they add to 1. I definitely think the answer for the question at the top is 2/9 and people are taking a method out of context.

Thanks, AlmostFamous. I am referring to the example problem under Combinatorics/Domino Effect in the advanced section in the Word Translations guide on Page 190.

I am confused as the two approaches seem different:

1. In the Veritas approach, we just calculate the probability-which is the same irrespective of the order and stop.

2. In the MGMAT approach, we calculate the probability and multiply it by the different ways this can be done.

3. A third approach, would be simply to take the combination approach - that is

3 gumballs can be chosen from 16 in 16C3 ways

1 blue from 7 blue gumballs can be chosen in 7C1 ways & 1 green from 5 green gumballs in 5C1 ways & 4 red gumballs in 4C1 ways and hence

probability that 1 blue, 1 green & 1 red - 7X5X4/(16C3) = 1/4

but this does not take the dependency i.e. P(A&B&C) = P(A) . Pa(B). Pab(C) as defined in the Veritas book

i.e calculating B taking into account A has occured and calculating C taking into account A & B have occured.

That is why I thought the approach from MGMAT (2 here) seems wrong.

Would really appreciate it if someone could point out which would be the correct way to go about solving problems of this type.

I basically studied all the concepts from the Veritas book and applied them to the MGMAT problems. My understanding is that this is the only conflicting concept - All others the answers correspond.

No. The answer to the question above is 1/4.
The machine dispenses the 3 gumballs, one after another.
It could do it in 6 different ways: BGR, GRB, BRG, etc

Explanation: The 'Veritas Method' (actually it is simultaneous events method) is used when it is mentioned 'three balls are dispensed simultaneously'. Here the questions says, 'three balls are dispensed. What is the probability that one is red, one green and one blue?' It does not say that the three balls are dispensed simultaneously.

There is a difference.

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?

Here I pick two balls. It doesn't say I pick them simultaneously. I can pick them in 6 ways:
1 R, 1 B
1B, 1 R
1 R, 1G
1G, 1R
1B, 1G
1G, 1B
Probability of each is 1/6. Probability of 1 red and 1 blue is 2*1/6 = 1/3 since there are 2 such cases.

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?

Now the order does not matter. The probability of 1/3 we got above is divided by 2 because 1/3 includes 2 cases: 1R, 1B and 1B, 1R.
So the probability here is 1/6. Same as the probabilty of 1 Red and then 1 Blue. Also same as probability of 1 Blue and then 1 Red.
It is counter intuitive. But mathematicians have proved it conclusively.