Total number of marbles = 8 + y

Probability of 2 red marbles = (8C2)/(8+y C 2)
Probability of 1 marble of each color = 2*8/8+y * y/7+y

1. y < = 8 Pick max value of 8
=> Probability of 2 red marbles = 8C2 / 16C2 = 28 / 120 = 7/30

=> Probability of 1 of each color = 2* 8/16 * 8/15 = 8/15

So it is more likely that you would have one of each color. Now consider y =1

=> Probability of 2 red marbles = 8C2 / 9C2 = 28 / 36 = 7/9

=> Probability of 1 of each color = 2 * 8/9 * 1/8 = 2/9

So here it is more likely to have 2 red marbles.

INSUFFICIENT.


2. Pick y = 4

=> Probability of 2 red marbles = 8C2 / 12C2 = 28 / 66 = 14/33

=> Probability of 1 of each color = 2* 8/12 * 4/11 = 16 /33

So it is more likely to have 1 of each color.

Increasing y above this will only increase the likelihood of getting 1 of each color, as can be observed by the values of the probabilities at y = 8.

The answer is therefore always - No
SUFFICIENT

Pick D. EDIT: Oops I meant B, thanks for catching that, outreach!


There may be an easier way to do this, but I would have just tested these 3 cases.

@albatross86
i think by mistake u picked D as the answer
B is correct

outreach wrote:@albatross86
i think by mistake u picked D as the answer
B is correct

Yeah sorry, I meant B

Probability is about # of things available of what you want divided by the total number of things. If you want one thing AND another thing you multiply those fractions. If you want one thing OR another thing you add those fractions. (there is also a way for doing exactly and at least but those don't apply here). this problem also deals with inequality when generally means solving the equations to check the values.

We have to determine if it is more likely (the probability is greater) that she will have 2 red marbles or one of each so we have to do two probabilities. 1 for a red AND red and 1 for one of each color (This is (R AND Y) or (Y AND R))

for two reds the probability is 8/(8+y) *7/(7+y) = 56/(8+y)(7+y)

for one of each color it is (8/(8+y) * (y/(7+y)) + {(y/(8+y) * 8/(7+y)} = 8y/(8+y)(7+y) + 8y/(8 +y)(7+y) =
16y/(8 +y)(7+y)

Because the denominators are the same we are really asked whether 56>16y

Stmt 1 says y<=8 which doesn't help us becuause that may lead to either greater than 56 or less than 56

if y >= 4 then it always works becuase even if it is at it's lowest number 16y=64 which is greater than 56.

Great explanations Becky and Albatross.

Great explanation Becky..

tpr-becky wrote: for two reds the probability is 8/(8+y) *7/(7+y) = 56/(8+y)(7+y)

I want to understand one thing.The probability formula that you have put up above holds good for case when reds are picked one after another without replacement. What if both reds are picked at a time.

Then, the formula will become 56 / (8+y)(7+y) / (2) = 56*2/(8+y)(7+y) [ Total selection = (8+Y)C2 ]

So, when question does not mention anything, how do we assume one or the other ?

Edited the typo..

sumanr84 wrote:Great explanation Becky..

tpr-becky wrote: for two reds the probability is 8/(8+y) *7/(7+y) = 56/(8+y)(7+y)

I want to understand one thing.The probability formula that you have put up above holds good for case when reds are picked one after another without replacement. What if both reds are picked at a time.

Then, the formula will become 56 / (8+y)(7+y) (2) [ Total selection = (8+Y)C2 ]

So, when question does not mention anything, how do we assume one or the other ?

Hi,

there's no mathematical difference between "one at a time without replacement" and "both simultaneously" - you can treat those two cases identically. Becky's application of the formula is correct in both cases (I'm not sure how you derived your formula, but it's incorrect if it yields a different result).

Thanks Stuart for the reply.

My prev post has a typo. Actually I mean that total selection of 2 items out of (8+Y) would be (8+Y) C2

Assume Y =2,

If I use my formula: The total selection available = 10C2 = 45

If I use Becky's formula : The total selection available = 10 * 9 = 90

@Suman - I see what you are saying, but actually you are making a slight error.

How did you get 10*9 from Becky's formula? Remember -she has not made some general formula here, but simply split up the two events of picking the marbles up. The denominator of her expression (8+y)*(7+y) is a simplified expression and does not denote the total number of selections as you seem to have assumed.

Coming back to your first post, why did you divide the denominator by 2?

When you say that you are picking both reds "at the same time" you are forgetting that this is not practically different from picking them up one at a time.

CASE 1: There are 10 balls. I pick 1. There are 9 left. I pick another out of these 9.

CASE 2: There are 10 balls. I take 2 out.

In CASE 2, is it possible that I can take the same ball out twice? No! So even though I am "lifting" both balls out simultaneously, I have actually made a decision to choose 1, and then choose the other.

I am not sure if this makes sense but it is a common mistake, and I completely agree with Stuart.

This can be proven by testing the following case.

For y = 2. With Becky's formula of 56/(8+y)(7+y), the probability of 2 red balls is simply 28/45

Using combinatorics:

Selections with red balls = 8C2 = 28
Total selections = 10C2 = 45

So Probability = 28/45

Same answer. So whether or not you treat the 2 balls as together or one at a time, real world tells us even if the event is happening at the same time you are still choosing 1 at a time since you cannot choose the same ball twice.

For more about this you can search the topic of "Dependent events in probability"

Hope this clarifies your doubt!

@albatross : Thanks for taking time in making it real simple.

I realized my mistake after my post. What I initially assumed is that Becky's formula is making total selection as (8+Y) * (7+Y), but then I realized that she is just multiplying individual probability of occurrence.
So, then I calculated prob using my formula and arrived at the same result as you have clearly mentioned.

You even cleared one more doubt of "Dependent events in probability" ..--Thanks