5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag

The probability that the result will be 2 blue and 2 purple marbles:
(5/12)*(5/12)*(4/12)*(4/12)=25/1296

The probability that it will not happen should be 1-(25/1296)=1271/1296.... and this is not an equivalent of D...

Where is my mistake?

No cigar yet Simaq!

Its without replacement, and above u have done the equivalent of with replacement. E.g whats the probability of picking 2 blue marbles out of a box containing 5 blue, 2 red, 3 black without replacement?

5/10 * 4/9 = 20/90 = 2/9

Ok missed the replacement part....

Then the long way:

blue blue purple purple: 5/12*4/11*4/10*3/9
purple purple blue blue: 4/12*3/11*5/10*4/10
purple blue purple blue: 4/12*5/11*3/10*5/9
blue purple blue purple: 5/12*4/11*4/10*3/9

1-sum of all those values... must be a shortcut though

How would this answer be different if it said "with replacement"? Can you provide the #of combinations in that case?

I'm thinking that with replacement you would have a total number of possible draws of 12^4 (could choose any of the 12 with any single draw). I think the numerator would look like 5*5*4*4 (5 ways to choose the first blue marble times 5 ways to choose the second blue marble times 4 ways to choose the first purple marble, etc.). Hence we would ultimately have

1 - 25*16/12^4 = 98% chance of not having this draw

Howver, after reading this post probability-colored-balls-55253.html#p637525 I'm confused. Can anyone chime in an clear up the correct way to compute the probability with replacement?

Excellent Bunuel. Now if you can clarify one more point. Same cases as 1 and 2 above, but in this case the question is - we pull balls 1by1 , what is the probability of seeing a blue one in the 9th draw (i) with replacement - i think this is easy 5/12 (ii) without replacement - i believe the answer here is also 5/12. My question how?

A jar contains 5 blue balls and 7 black balls. One by one, every ball is selected at random without replacement. What is the probability that the 9th ball selected is blue?

The probability of drawing blue ball is 5/12 and it will not change for ANY successive drawing: second, third, fourth...

Obviously for case with replacement the answer would be the same - every time we are drawing from 12 balls out of which 5 are blue, so every time probability of drawing blue ball is 5/12.

Similar problems:
probability-of-picking-last-97015.html?hilit=initial%20probability
probability-question-please-solve-90272.html?hilit=initial%20probability#p686028

Hope it helps.

thanks a lot Bunuel

Yes excellent. With or without replacement gives same answer. Can you give me an example where
The results are different?

Also while in the topic of probability can you answer the following please, my guesses are mentioned

Distribute 10 different rocks among 15 people ? 10^15
Distribute 10 identical rocks among 15 ? Not sure
Distribute 15 different rocks among 10 ? 15^10
Distribute 15 identical rocks among 10? 15+9 C 9
thanks

Posted from my mobile device

Posted from my mobile device

Distribute 10 different rocks among 15 people? 10^15
Distribute 10 identical rocks among 15? (10+15-1)C(15-1)=24C14
Distribute 15 different rocks among 10? 15^10
Distribute 15 identical rocks among 10? (15+10-1)C(10-1)=24C9

Direct formulas for identical objects:
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

I doubt you'll see division of identical object involving such a big numbers on GMAT or division of different objects when each can get at least n objects (the question you've asked before) involving such a big numbers. With smaller numbers you can solve with different approaches depending on problem. So I would not worry about this questions too much. Better to concentrate on other areas.

--> WHY??

If I draw 1 blue ball only 4 will be left (no replacement...) - next prob. of drawing blue should be 4/11 ??

I think you misinterpreted the question (or I made it ambiguous). The question is: what is the probability that the 9th ball will be blue and it's meant that we don't know the results of first 8 draws (or drawing has not started yet and we are just curious what is the probability that 9th ball will be blue). If I pick one by one and every time I know the result then of course the probability of drawing of a blue ball will change depending on results of previous drawings.

Consider the following: these 12 balls were put in a line randomly. Now, what is the probability that the 9th ball will be blue? ANY ball in a line has equal chances to be blue and ANY ball has equal chances to be black. Why would the probability of say 4th and 7th balls of being blue be different? For any ball in a line p=5/12 to be blue and for any ball p=7/12 to be black: for any ball in a line the probability of being blue plus the probability of being black must be 1, as there are only black and blue balls in a line. This example is basically the same as our original question, so answer for original question is also 5/12.

Follow the links in my previous post for similar problems.

Hope it helps.

Makes more sense now, thanks! Was just relatively hard to comprehend at first, what really was meant, but you made it clear!

Thanks Bunuel. I just chose the numbers at. Random, I see GMAT use smaller numbers, and then some start existing options, but the formula is more powerful. Also was curious to know how identical vs different changed the answers.

I wouldn't have known the replacement vs nonreplacement giving the same probability, thanks for the explanation to the prioranswer.

Posted from my mobile device

Is there a way to solve WITH replacement cases using combinatorics? I notice that all WITH replacement cases have only been solved using only one method.

Thank you Bunuel !
I wish you had been there in my school for maths

Posted from my mobile device

We can use the formula:

P(will not be 2 blue and 2 purple marbles) = 1 - P(2 blue and 2 purple marbles)

We can select two blue marbles in 5C2 = (5 x 4)/2! = 10 ways.

We can select 2 purple marbles in 4C2 = (4 x 3)/2! = 6 ways.

Thus, the total number of pairings of 2 blue and 2 purple marbles is 10 x 6 = 60 ways.

We can select 4 marbles from 12 in 12C4 = (12 x 11 x 10 x 9)/(4 x 3 x 2) = 11 x 5 x 9 = 495 ways.

Thus, P(2 blue and 2 purple marbles) = 60/495 = 20/165 = 4/33.

So, P(will not be 2 blue and 2 purple marbles) = 1 - 4/33 = 29/33.

Answer: E