Rachel is throwing a die with sides numbered consecutively

A for me too.

(1) says dice has 5 sides.
probability of NOT gettign ANY x in two throws: 4/5 * 4/5 = 16/25
therefore, probability of getting ATLEAST one x = 1 - 16/25 = 9/25

(2) 61/25 > 1. Can we have probabilities greater than 1???

Well, it's not A, so it has to be D. btw Macedon, I assume you meant to type 16/25 and not 61/25. Even so, one more adjustment is needed: I think the second statement should read 9/25 and not 16/25. Otherwise, I don't see how this can be solved.
-----------------
Statement 1: the probability equals
1 - (4/5)^2 = 9/25, where 4/5 is the probability not to get an x

Statement 2: similar reasoning as above. The probability to get at least 1 x equals 1-(the probability to get no x's). On every throw the probability not to get an x equals (x-1)/x. Let's say we throw the die y times, so we have

P(1 or more x) = 1 - ((x-1)/x)^y

and 1 - ((x-1)/x)^(y+1) = 9/25

=> ((x-1)/x)^(y+1) = 16/25

x-1 and x are consecutive numbers. For any power n, there is no way that (x-1/x)^n can be presented as (simplified to) anything other than ((x-1)^n)/(x^n), because the two numbers do not have any common factors. I mean that 16/25 must represent the second power of 4 divided by the second power of 5.

We can therefore say that x=5 and y=1. The statement is sufficient.

I figured B is obviously not sufficient alone because it say "one more time" and we do not know one more time from what many number of times?

Ranga, I can't come up with a better explanation than what I put up above. The validity of my solution is contingent upon a wrong stem.

If thе text of the problem is exactly as in Macedon's first post, I agree with A.

THIS IS A QUESTION FROM A CAT OF PRINCETON WEBSITE....HERE IT IS THE EXPLANATION...

Yes. The pieces of the puzzle approach is useful in solving this problem. To find the probability of getting at least one x, you need to know the number of faces on the die (x) and the number of times the die is thrown. Statement (1) gives the number of times the die is thrown and also means that x = 5. The correct answer is either A or D. Statement (2) is also sufficient. To find the probability of getting at least one x, you would find the probability of not getting an x on any of the throws and subtract that probability from 1. To find the probability of not getting an x on any of the throws would require multiplying the same fraction together some number of times. Since 125^1/3= 5, the die was tossed three times in Statement (2) and it had 5 faces. The correct answer is D.


AS I SAID I PICKED A TOO..CIAO A TUTTI

I assume that the actual question stated on (2) 61/125

1-[(x-1)/x]^(y+1) = 61/125
[(x-1)/x]^(y+1) = 64/125 = (4/5)^3 ===> x=5 ; y=2.

Bunuel, Your Intervention Please. This is a good question.

Did not get it confused ,, Bunuel please help to better understand this

First of all, this is a hard question from advanced topic, so it should be attempted only when all fundamentals are clear and there is no problem whatsoever in solving easier questions. Next, you really need to be more specific when asking a question.

Bunuel : I got the 1st statement case where given is that the P(getting an x in each throw) =1/5 So, P(no X in each throw)=1- 1/5=4/5 and P(Atleast 1x)=1-P(No x)in each throw = 1- 4/5*4/5.

But in 2nd : I am lost at the point where you explained : "Since P(no x's) for n throws also equals to (x-1/x)^n, then ........"

The probability of getting x is 1/x --> the probability of not getting x is 1 - 1/x = (x-1)/x. The probability of not getting x when throwing n times is therefore \((\frac{x-1}{x})^n\).

Bunuel

The question says that probability of getting x on either throw is 1/5.

Shouldn't this mean P(getting x at throw 1) + P( getting X at throw 2) = 1/5

I.e. 1/x + (x-1)/x^2 =1/5?

Posted from my mobile device

This actually triggers a basic question. The probability of getting a 6 on either throw of a dice, given it's thrown twice should be 1/6 + 5/36 = 11/36 i.e slightly less than 1/3. Why are we assuming the probability of X being shown on either throw to be x=5.


Please guys help!!!

Posted from my mobile device

Bunuel chetan2u

If you have a moment, I just wanted to clarify that I am reading the question correctly.

In statement 1, when we are told “the probability of getting an X on EITHER of the two times” —-/> is the meaning as follows?

Because they are independent events, any time she throws the die, she has a 1 in 5 chance of X showing up. In other words, any time she throws the die the Probability of an X showing up = (1/5)

I initially interpreted the meaning as follows:

P of getting an X on either of the 2 times =

(P of getting X on 1st throw AND getting any other number on 2nd throw)

+

(P of getting any other number on 1st throw AND getting X on 2nd throw)

+

(P of getting X on 1st throw AND X on 2nd throw)

I believe I wasn’t reading the statement close enough. The singular “die” is used. Therefore, I should be understanding the meaning as:

For any independent throw that she makes, she has a (1/5) chance of getting the X to show up.

Am I interpreting the statement correctly? Just looking to clarify and any help would be appreciated. Thank you.

Edit: and I missed the official explanation from Princeton review that someone posted.

There is only one X # on one face, so the Probability of X showing up must have a 1 in the Numerator.

If the probability of getting an x when she “throws the DIE” is = (1/5)

Then there must be 5 consecutive numbers on the die.






Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.